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TWEET # Project Euler Answers < 200000000 a guest Aug 2nd, 2015 4,702 Never
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2.
3.
4. Project Euler is protected under
5. Attribution-Non-Commercial-Share Alike 2.0 UK: England & Wales
6.
7. ^sup and [sub] are frequently use for subscripts/superscripts.
8. Many symbols are utf8, my apologies if you are on a 7-bit TTY.
9.
10. Solutions are hashed with md5sum
11. echo -n 'myanswer' | md5sum
12.
13.
14.
15. generated on 2014-6-29 at 15:46
16.
17.
18.
19. Problem 1
20. =========
21.
22.
23.    If we list all the natural numbers below 10 that are multiples of 3 or 5,
24.    we get 3, 5, 6 and 9. The sum of these multiples is 23.
25.
26.    Find the sum of all the multiples of 3 or 5 below 1000.
27.
28.
30.
31.
32. Problem 2
33. =========
34.
35.
36.    Each new term in the Fibonacci sequence is generated by adding the
37.    previous two terms. By starting with 1 and 2, the first 10 terms will be:
38.
39.                      1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
40.
41.    By considering the terms in the Fibonacci sequence whose values do not
42.    exceed four million, find the sum of the even-valued terms.
43.
44.
46.
47.
48. Problem 3
49. =========
50.
51.
52.    The prime factors of 13195 are 5, 7, 13 and 29.
53.
54.    What is the largest prime factor of the number 600851475143 ?
55.
56.
58.
59.
60. Problem 4
61. =========
62.
63.
64.    A palindromic number reads the same both ways. The largest palindrome made
65.    from the product of two 2-digit numbers is 9009 = 91 × 99.
66.
67.    Find the largest palindrome made from the product of two 3-digit numbers.
68.
69.
71.
72.
73. Problem 5
74. =========
75.
76.
77.    2520 is the smallest number that can be divided by each of the numbers
78.    from 1 to 10 without any remainder.
79.
80.    What is the smallest positive number that is evenly divisible by all of
81.    the numbers from 1 to 20?
82.
83.
85.
86.
87. Problem 6
88. =========
89.
90.
91.    The sum of the squares of the first ten natural numbers is,
92.
93.                           1^2 + 2^2 + ... + 10^2 = 385
94.
95.    The square of the sum of the first ten natural numbers is,
96.
97.                        (1 + 2 + ... + 10)^2 = 55^2 = 3025
98.
99.    Hence the difference between the sum of the squares of the first ten
100.    natural numbers and the square of the sum is 3025 − 385 = 2640.
101.
102.    Find the difference between the sum of the squares of the first one
103.    hundred natural numbers and the square of the sum.
104.
105.
107.
108.
109. Problem 7
110. =========
111.
112.
113.    By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
114.    that the 6th prime is 13.
115.
116.    What is the 10 001st prime number?
117.
118.
120.
121.
122. Problem 8
123. =========
124.
125.
126.    The four adjacent digits in the 1000-digit number that have the greatest
127.    product are 9 × 9 × 8 × 9 = 5832.
128.
129.                73167176531330624919225119674426574742355349194934
130.                96983520312774506326239578318016984801869478851843
131.                85861560789112949495459501737958331952853208805511
132.                12540698747158523863050715693290963295227443043557
133.                66896648950445244523161731856403098711121722383113
134.                62229893423380308135336276614282806444486645238749
135.                30358907296290491560440772390713810515859307960866
136.                70172427121883998797908792274921901699720888093776
137.                65727333001053367881220235421809751254540594752243
138.                52584907711670556013604839586446706324415722155397
139.                53697817977846174064955149290862569321978468622482
140.                83972241375657056057490261407972968652414535100474
141.                82166370484403199890008895243450658541227588666881
142.                16427171479924442928230863465674813919123162824586
143.                17866458359124566529476545682848912883142607690042
144.                24219022671055626321111109370544217506941658960408
145.                07198403850962455444362981230987879927244284909188
146.                84580156166097919133875499200524063689912560717606
147.                05886116467109405077541002256983155200055935729725
148.                71636269561882670428252483600823257530420752963450
149.
150.    Find the thirteen adjacent digits in the 1000-digit number that have the
151.    greatest product. What is the value of this product?
152.
153.
155.
156.
157. Problem 9
158. =========
159.
160.
161.    A Pythagorean triplet is a set of three natural numbers, a < b < c, for
162.    which,
163.
164.                                 a^2 + b^2 = c^2
165.
166.    For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
167.
168.    There exists exactly one Pythagorean triplet for which a + b + c = 1000.
169.    Find the product abc.
170.
171.
173.
174.
175. Problem 10
176. ==========
177.
178.
179.    The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
180.
181.    Find the sum of all the primes below two million.
182.
183.
185.
186.
187. Problem 11
188. ==========
189.
190.
191.    In the 20×20 grid below, four numbers along a diagonal line have been
192.    marked in red.
193.
194.           08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
195.           49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
196.           81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
197.           52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
198.           22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
199.           24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
200.           32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
201.           67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
202.           24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
203.           21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
204.           78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
205.           16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
206.           86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
207.           19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
208.           04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
209.           88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
210.           04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
211.           20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
212.           20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
213.           01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
214.
215.    The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
216.
217.    What is the greatest product of four adjacent numbers in the same
218.    direction (up, down, left, right, or diagonally) in the 20×20 grid?
219.
220.
222.
223.
224. Problem 12
225. ==========
226.
227.
228.    The sequence of triangle numbers is generated by adding the natural
229.    numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 =
230.    28. The first ten terms would be:
231.
232.                     1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
233.
234.    Let us list the factors of the first seven triangle numbers:
235.
236.       1: 1
237.       3: 1,3
238.       6: 1,2,3,6
239.      10: 1,2,5,10
240.      15: 1,3,5,15
241.      21: 1,3,7,21
242.      28: 1,2,4,7,14,28
243.
244.    We can see that 28 is the first triangle number to have over five
245.    divisors.
246.
247.    What is the value of the first triangle number to have over five hundred
248.    divisors?
249.
250.
252.
253.
254. Problem 13
255. ==========
256.
257.
258.    Work out the first ten digits of the sum of the following one-hundred
259.    50-digit numbers.
260.
261.                37107287533902102798797998220837590246510135740250
262.                46376937677490009712648124896970078050417018260538
263.                74324986199524741059474233309513058123726617309629
264.                91942213363574161572522430563301811072406154908250
265.                23067588207539346171171980310421047513778063246676
266.                89261670696623633820136378418383684178734361726757
267.                28112879812849979408065481931592621691275889832738
268.                44274228917432520321923589422876796487670272189318
269.                47451445736001306439091167216856844588711603153276
270.                70386486105843025439939619828917593665686757934951
271.                62176457141856560629502157223196586755079324193331
272.                64906352462741904929101432445813822663347944758178
273.                92575867718337217661963751590579239728245598838407
274.                58203565325359399008402633568948830189458628227828
275.                80181199384826282014278194139940567587151170094390
276.                35398664372827112653829987240784473053190104293586
277.                86515506006295864861532075273371959191420517255829
278.                71693888707715466499115593487603532921714970056938
279.                54370070576826684624621495650076471787294438377604
280.                53282654108756828443191190634694037855217779295145
281.                36123272525000296071075082563815656710885258350721
282.                45876576172410976447339110607218265236877223636045
283.                17423706905851860660448207621209813287860733969412
284.                81142660418086830619328460811191061556940512689692
285.                51934325451728388641918047049293215058642563049483
286.                62467221648435076201727918039944693004732956340691
287.                15732444386908125794514089057706229429197107928209
288.                55037687525678773091862540744969844508330393682126
289.                18336384825330154686196124348767681297534375946515
290.                80386287592878490201521685554828717201219257766954
291.                78182833757993103614740356856449095527097864797581
292.                16726320100436897842553539920931837441497806860984
293.                48403098129077791799088218795327364475675590848030
294.                87086987551392711854517078544161852424320693150332
295.                59959406895756536782107074926966537676326235447210
296.                69793950679652694742597709739166693763042633987085
297.                41052684708299085211399427365734116182760315001271
298.                65378607361501080857009149939512557028198746004375
299.                35829035317434717326932123578154982629742552737307
300.                94953759765105305946966067683156574377167401875275
301.                88902802571733229619176668713819931811048770190271
302.                25267680276078003013678680992525463401061632866526
303.                36270218540497705585629946580636237993140746255962
304.                24074486908231174977792365466257246923322810917141
305.                91430288197103288597806669760892938638285025333403
306.                34413065578016127815921815005561868836468420090470
307.                23053081172816430487623791969842487255036638784583
308.                11487696932154902810424020138335124462181441773470
309.                63783299490636259666498587618221225225512486764533
310.                67720186971698544312419572409913959008952310058822
311.                95548255300263520781532296796249481641953868218774
312.                76085327132285723110424803456124867697064507995236
313.                37774242535411291684276865538926205024910326572967
314.                23701913275725675285653248258265463092207058596522
315.                29798860272258331913126375147341994889534765745501
316.                18495701454879288984856827726077713721403798879715
317.                38298203783031473527721580348144513491373226651381
318.                34829543829199918180278916522431027392251122869539
319.                40957953066405232632538044100059654939159879593635
320.                29746152185502371307642255121183693803580388584903
321.                41698116222072977186158236678424689157993532961922
322.                62467957194401269043877107275048102390895523597457
323.                23189706772547915061505504953922979530901129967519
324.                86188088225875314529584099251203829009407770775672
325.                11306739708304724483816533873502340845647058077308
326.                82959174767140363198008187129011875491310547126581
327.                97623331044818386269515456334926366572897563400500
328.                42846280183517070527831839425882145521227251250327
329.                55121603546981200581762165212827652751691296897789
330.                32238195734329339946437501907836945765883352399886
331.                75506164965184775180738168837861091527357929701337
332.                62177842752192623401942399639168044983993173312731
333.                32924185707147349566916674687634660915035914677504
334.                99518671430235219628894890102423325116913619626622
335.                73267460800591547471830798392868535206946944540724
336.                76841822524674417161514036427982273348055556214818
337.                97142617910342598647204516893989422179826088076852
338.                87783646182799346313767754307809363333018982642090
339.                10848802521674670883215120185883543223812876952786
340.                71329612474782464538636993009049310363619763878039
341.                62184073572399794223406235393808339651327408011116
342.                66627891981488087797941876876144230030984490851411
343.                60661826293682836764744779239180335110989069790714
344.                85786944089552990653640447425576083659976645795096
345.                66024396409905389607120198219976047599490197230297
346.                64913982680032973156037120041377903785566085089252
347.                16730939319872750275468906903707539413042652315011
348.                94809377245048795150954100921645863754710598436791
349.                78639167021187492431995700641917969777599028300699
350.                15368713711936614952811305876380278410754449733078
351.                40789923115535562561142322423255033685442488917353
352.                44889911501440648020369068063960672322193204149535
353.                41503128880339536053299340368006977710650566631954
354.                81234880673210146739058568557934581403627822703280
355.                82616570773948327592232845941706525094512325230608
356.                22918802058777319719839450180888072429661980811197
357.                77158542502016545090413245809786882778948721859617
358.                72107838435069186155435662884062257473692284509516
359.                20849603980134001723930671666823555245252804609722
360.                53503534226472524250874054075591789781264330331690
361.
363.
364.
365. Problem 14
366. ==========
367.
368.
369.    The following iterative sequence is defined for the set of positive
370.    integers:
371.
372.    n → n/2 (n is even)
373.    n → 3n + 1 (n is odd)
374.
375.    Using the rule above and starting with 13, we generate the following
376.    sequence:
377.
378.                    13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
379.
380.    It can be seen that this sequence (starting at 13 and finishing at 1)
381.    contains 10 terms. Although it has not been proved yet (Collatz Problem),
382.    it is thought that all starting numbers finish at 1.
383.
384.    Which starting number, under one million, produces the longest chain?
385.
386.    NOTE: Once the chain starts the terms are allowed to go above one million.
387.
388.
390.
391.
392. Problem 15
393. ==========
394.
395.
396.    Starting in the top left corner of a 2×2 grid, and only being able to move
397.    to the right and down, there are exactly 6 routes to the bottom right
398.    corner.
399.
400.    How many such routes are there through a 20×20 grid?
401.
402.
403.    p_015.gif
405.
406.
407. Problem 16
408. ==========
409.
410.
411.    2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
412.
413.    What is the sum of the digits of the number 2^1000?
414.
415.
417.
418.
419. Problem 17
420. ==========
421.
422.
423.    If the numbers 1 to 5 are written out in words: one, two, three, four,
424.    five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
425.
426.    If all the numbers from 1 to 1000 (one thousand) inclusive were written
427.    out in words, how many letters would be used?
428.
429.    NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
430.    forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
431.    20 letters. The use of "and" when writing out numbers is in compliance
432.    with British usage.
433.
434.
436.
437.
438. Problem 18
439. ==========
440.
441.
442.    By starting at the top of the triangle below and moving to adjacent
443.    numbers on the row below, the maximum total from top to bottom is 23.
444.
445.                                        3
446.                                       7 4
447.                                      2 4 6
448.                                     8 5 9 3
449.
450.    That is, 3 + 7 + 4 + 9 = 23.
451.
452.    Find the maximum total from top to bottom of the triangle below:
453.
454.                                        75
455.                                      95 64
456.                                     17 47 82
457.                                   18 35 87 10
458.                                  20 04 82 47 65
459.                                19 01 23 75 03 34
460.                               88 02 77 73 07 63 67
461.                             99 65 04 28 06 16 70 92
462.                            41 41 26 56 83 40 80 70 33
463.                          41 48 72 33 47 32 37 16 94 29
464.                         53 71 44 65 25 43 91 52 97 51 14
465.                       70 11 33 28 77 73 17 78 39 68 17 57
466.                      91 71 52 38 17 14 91 43 58 50 27 29 48
467.                    63 66 04 68 89 53 67 30 73 16 69 87 40 31
468.                   04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
469.
470.    NOTE: As there are only 16384 routes, it is possible to solve this problem
471.    by trying every route. However, Problem 67, is the same challenge with
472.    a triangle containing one-hundred rows; it cannot be solved by brute
473.    force, and requires a clever method! ;o)
474.
475.
477.    1. problem=67
479.
480.
481. Problem 19
482. ==========
483.
484.
485.    You are given the following information, but you may prefer to do some
486.    research for yourself.
487.
488.      • 1 Jan 1900 was a Monday.
489.      • Thirty days has September,
490.        April, June and November.
491.        All the rest have thirty-one,
492.        Saving February alone,
493.        Which has twenty-eight, rain or shine.
494.        And on leap years, twenty-nine.
495.      • A leap year occurs on any year evenly divisible by 4, but not on a
496.        century unless it is divisible by 400.
497.
498.    How many Sundays fell on the first of the month during the twentieth
499.    century (1 Jan 1901 to 31 Dec 2000)?
500.
501.
503.
504.
505. Problem 20
506. ==========
507.
508.
509.    n! means n × (n − 1) × ... × 3 × 2 × 1
510.
511.    For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
512.    and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 =
513.    27.
514.
515.    Find the sum of the digits in the number 100!
516.
517.
519.
520.
521. Problem 21
522. ==========
523.
524.
525.    Let d(n) be defined as the sum of proper divisors of n (numbers less than
526.    n which divide evenly into n).
527.    If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair
528.    and each of a and b are called amicable numbers.
529.
530.    For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
531.    44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1,
532.    2, 4, 71 and 142; so d(284) = 220.
533.
534.    Evaluate the sum of all the amicable numbers under 10000.
535.
536.
538.
539.
540. Problem 22
541. ==========
542.
543.
544.    Using names.txt, a 46K text file containing over five-thousand first
545.    names, begin by sorting it into alphabetical order. Then working out the
546.    alphabetical value for each name, multiply this value by its alphabetical
547.    position in the list to obtain a name score.
548.
549.    For example, when the list is sorted into alphabetical order, COLIN, which
550.    is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So,
551.    COLIN would obtain a score of 938 × 53 = 49714.
552.
553.    What is the total of all the name scores in the file?
554.
555.
557.    1. names.txt
559.
560.
561. Problem 23
562. ==========
563.
564.
565.    A perfect number is a number for which the sum of its proper divisors is
566.    exactly equal to the number. For example, the sum of the proper divisors
567.    of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
568.    number.
569.
570.    A number n is called deficient if the sum of its proper divisors is less
571.    than n and it is called abundant if this sum exceeds n.
572.
573.    As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the
574.    smallest number that can be written as the sum of two abundant numbers is
575.    24. By mathematical analysis, it can be shown that all integers greater
576.    than 28123 can be written as the sum of two abundant numbers. However,
577.    this upper limit cannot be reduced any further by analysis even though it
578.    is known that the greatest number that cannot be expressed as the sum of
579.    two abundant numbers is less than this limit.
580.
581.    Find the sum of all the positive integers which cannot be written as the
582.    sum of two abundant numbers.
583.
584.
586.
587.
588. Problem 24
589. ==========
590.
591.
592.    A permutation is an ordered arrangement of objects. For example, 3124 is
593.    one possible permutation of the digits 1, 2, 3 and 4. If all of the
594.    permutations are listed numerically or alphabetically, we call it
595.    lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
596.
597.                        012   021   102   120   201   210
598.
599.    What is the millionth lexicographic permutation of the digits 0, 1, 2, 3,
600.    4, 5, 6, 7, 8 and 9?
601.
602.
604.
605.
606. Problem 25
607. ==========
608.
609.
610.    The Fibonacci sequence is defined by the recurrence relation:
611.
612.      F[n] = F[n−1] + F[n−2], where F = 1 and F = 1.
613.
614.    Hence the first 12 terms will be:
615.
616.      F = 1
617.      F = 1
618.      F = 2
619.      F = 3
620.      F = 5
621.      F = 8
622.      F = 13
623.      F = 21
624.      F = 34
625.      F = 55
626.      F = 89
627.      F = 144
628.
629.    The 12th term, F, is the first term to contain three digits.
630.
631.    What is the first term in the Fibonacci sequence to contain 1000 digits?
632.
633.
635.
636.
637. Problem 26
638. ==========
639.
640.
641.    A unit fraction contains 1 in the numerator. The decimal representation of
642.    the unit fractions with denominators 2 to 10 are given:
643.
644.      1/2  =  0.5
645.      1/3  =  0.(3)
646.      1/4  =  0.25
647.      1/5  =  0.2
648.      1/6  =  0.1(6)
649.      1/7  =  0.(142857)
650.      1/8  =  0.125
651.      1/9  =  0.(1)
652.      1/10 =  0.1
653.
654.    Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can
655.    be seen that 1/7 has a 6-digit recurring cycle.
656.
657.    Find the value of d < 1000 for which ^1/[d] contains the longest recurring
658.    cycle in its decimal fraction part.
659.
660.
662.
663.
664. Problem 27
665. ==========
666.
667.
668.    Euler discovered the remarkable quadratic formula:
669.
670.                                   n² + n + 41
671.
672.    It turns out that the formula will produce 40 primes for the consecutive
673.    values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41
674.    is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly
675.    divisible by 41.
676.
677.    The incredible formula  n² − 79n + 1601 was discovered, which produces 80
678.    primes for the consecutive values n = 0 to 79. The product of the
679.    coefficients, −79 and 1601, is −126479.
680.
681.    Considering quadratics of the form:
682.
683.      n² + an + b, where |a| < 1000 and |b| < 1000
684.
685.      where |n| is the modulus/absolute value of n
686.      e.g. |11| = 11 and |−4| = 4
687.
688.    Find the product of the coefficients, a and b, for the quadratic
689.    expression that produces the maximum number of primes for consecutive
690.    values of n, starting with n = 0.
691.
692.
694.
695.
696. Problem 28
697. ==========
698.
699.
700.    Starting with the number 1 and moving to the right in a clockwise
701.    direction a 5 by 5 spiral is formed as follows:
702.
703.                                  21 22 23 24 25
704.                                  20  7  8  9 10
705.                                  19  6  1  2 11
706.                                  18  5  4  3 12
707.                                  17 16 15 14 13
708.
709.    It can be verified that the sum of the numbers on the diagonals is 101.
710.
711.    What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral
712.    formed in the same way?
713.
714.
716.
717.
718. Problem 29
719. ==========
720.
721.
722.    Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
723.
724.      2^2=4, 2^3=8, 2^4=16, 2^5=32
725.      3^2=9, 3^3=27, 3^4=81, 3^5=243
726.      4^2=16, 4^3=64, 4^4=256, 4^5=1024
727.      5^2=25, 5^3=125, 5^4=625, 5^5=3125
728.
729.    If they are then placed in numerical order, with any repeats removed, we
730.    get the following sequence of 15 distinct terms:
731.
732.         4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
733.
734.    How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤
735.    100 and 2 ≤ b ≤ 100?
736.
737.
739.
740.
741. Problem 30
742. ==========
743.
744.
745.    Surprisingly there are only three numbers that can be written as the sum
746.    of fourth powers of their digits:
747.
748.      1634 = 1^4 + 6^4 + 3^4 + 4^4
749.      8208 = 8^4 + 2^4 + 0^4 + 8^4
750.      9474 = 9^4 + 4^4 + 7^4 + 4^4
751.
752.    As 1 = 1^4 is not a sum it is not included.
753.
754.    The sum of these numbers is 1634 + 8208 + 9474 = 19316.
755.
756.    Find the sum of all the numbers that can be written as the sum of fifth
757.    powers of their digits.
758.
759.
761.
762.
763. Problem 31
764. ==========
765.
766.
767.    In England the currency is made up of pound, £, and pence, p, and there
768.    are eight coins in general circulation:
769.
770.      1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
771.
772.    It is possible to make £2 in the following way:
773.
774.      1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
775.
776.    How many different ways can £2 be made using any number of coins?
777.
778.
780.
781.
782. Problem 32
783. ==========
784.
785.
786.    We shall say that an n-digit number is pandigital if it makes use of all
787.    the digits 1 to n exactly once; for example, the 5-digit number, 15234, is
788.    1 through 5 pandigital.
789.
790.    The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
791.    multiplicand, multiplier, and product is 1 through 9 pandigital.
792.
794.    identity can be written as a 1 through 9 pandigital.
795.
796.    HINT: Some products can be obtained in more than one way so be sure to
797.    only include it once in your sum.
798.
800.
801.
802. Problem 33
803. ==========
804.
805.
806.    The fraction 49/98 is a curious fraction, as an inexperienced
807.    mathematician in attempting to simplify it may incorrectly believe that
808.    49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
809.
810.    We shall consider fractions like, 30/50 = 3/5, to be trivial
811.    examples.
812.
813.    There are exactly four non-trivial examples of this type of fraction, less
814.    than one in value, and containing two digits in the numerator and
815.    denominator.
816.
817.    If the product of these four fractions is given in its lowest common
818.    terms, find the value of the denominator.
819.
820.
822.
823.
824. Problem 34
825. ==========
826.
827.
828.    145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
829.
830.    Find the sum of all numbers which are equal to the sum of the factorial of
831.    their digits.
832.
833.    Note: as 1! = 1 and 2! = 2 are not sums they are not included.
834.
835.
837.
838.
839. Problem 35
840. ==========
841.
842.
843.    The number, 197, is called a circular prime because all rotations of the
844.    digits: 197, 971, and 719, are themselves prime.
845.
846.    There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37,
847.    71, 73, 79, and 97.
848.
849.    How many circular primes are there below one million?
850.
851.
853.
854.
855. Problem 36
856. ==========
857.
858.
859.    The decimal number, 585 = 1001001001 (binary), is palindromic in both
860.    bases.
861.
862.    Find the sum of all numbers, less than one million, which are palindromic
863.    in base 10 and base 2.
864.
865.    (Please note that the palindromic number, in either base, may not include
867.
868.
870.
871.
872. Problem 37
873. ==========
874.
875.
876.    The number 3797 has an interesting property. Being prime itself, it is
877.    possible to continuously remove digits from left to right, and remain
878.    prime at each stage: 3797, 797, 97, and 7. Similarly we can work from
879.    right to left: 3797, 379, 37, and 3.
880.
881.    Find the sum of the only eleven primes that are both truncatable from left
882.    to right and right to left.
883.
884.    NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
885.
886.
888.
889.
890. Problem 38
891. ==========
892.
893.
894.    Take the number 192 and multiply it by each of 1, 2, and 3:
895.
896.      192 × 1 = 192
897.      192 × 2 = 384
898.      192 × 3 = 576
899.
900.    By concatenating each product we get the 1 to 9 pandigital, 192384576. We
901.    will call 192384576 the concatenated product of 192 and (1,2,3)
902.
903.    The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4,
904.    and 5, giving the pandigital, 918273645, which is the concatenated product
905.    of 9 and (1,2,3,4,5).
906.
907.    What is the largest 1 to 9 pandigital 9-digit number that can be formed as
908.    the concatenated product of an integer with (1,2, ... , n) where n > 1?
909.
910.
912.
913.
914. Problem 39
915. ==========
916.
917.
918.    If p is the perimeter of a right angle triangle with integral length
919.    sides, {a,b,c}, there are exactly three solutions for p = 120.
920.
921.    {20,48,52}, {24,45,51}, {30,40,50}
922.
923.    For which value of p ≤ 1000, is the number of solutions maximised?
924.
925.
927.
928.
929. Problem 40
930. ==========
931.
932.
933.    An irrational decimal fraction is created by concatenating the positive
934.    integers:
935.
936.                      0.123456789101112131415161718192021...
937.
938.    It can be seen that the 12^th digit of the fractional part is 1.
939.
940.    If d[n] represents the n^th digit of the fractional part, find the value
941.    of the following expression.
942.
943.       d × d × d × d × d × d × d
944.
945.
947.
948.
949. Problem 41
950. ==========
951.
952.
953.    We shall say that an n-digit number is pandigital if it makes use of all
954.    the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital
955.    and is also prime.
956.
957.    What is the largest n-digit pandigital prime that exists?
958.
959.
961.
962.
963. Problem 42
964. ==========
965.
966.
967.    The n^th term of the sequence of triangle numbers is given by, t[n] =
968.    ½n(n+1); so the first ten triangle numbers are:
969.
970.                     1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
971.
972.    By converting each letter in a word to a number corresponding to its
973.    alphabetical position and adding these values we form a word value. For
974.    example, the word value for SKY is 19 + 11 + 25 = 55 = t. If the word
975.    value is a triangle number then we shall call the word a triangle word.
976.
977.    Using words.txt, a 16K text file containing nearly two-thousand common
978.    English words, how many are triangle words?
979.
980.
982.    1. words.txt
984.
985.
986. Problem 43
987. ==========
988.
989.
990.    The number, 1406357289, is a 0 to 9 pandigital number because it is made
991.    up of each of the digits 0 to 9 in some order, but it also has a rather
992.    interesting sub-string divisibility property.
993.
994.    Let d be the 1^st digit, d be the 2^nd digit, and so on. In this
995.    way, we note the following:
996.
997.      • ddd=406 is divisible by 2
998.      • ddd=063 is divisible by 3
999.      • ddd=635 is divisible by 5
1000.      • ddd=357 is divisible by 7
1001.      • ddd=572 is divisible by 11
1002.      • ddd=728 is divisible by 13
1003.      • ddd=289 is divisible by 17
1004.
1005.    Find the sum of all 0 to 9 pandigital numbers with this property.
1006.
1007.
1009.
1010.
1011. Problem 44
1012. ==========
1013.
1014.
1015.    Pentagonal numbers are generated by the formula, P[n]=n(3n−1)/2. The first
1016.    ten pentagonal numbers are:
1017.
1018.                   1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
1019.
1020.    It can be seen that P + P = 22 + 70 = 92 = P. However, their
1021.    difference, 70 − 22 = 48, is not pentagonal.
1022.
1023.    Find the pair of pentagonal numbers, P[j] and P[k], for which their sum
1024.    and difference are pentagonal and D = |P[k] − P[j]| is minimised; what is
1025.    the value of D?
1026.
1027.
1029.
1030.
1031. Problem 45
1032. ==========
1033.
1034.
1035.    Triangle, pentagonal, and hexagonal numbers are generated by the following
1036.    formulae:
1037.
1038.    Triangle     T[n]=n(n+1)/2    1, 3, 6, 10, 15, ...
1039.    Pentagonal   P[n]=n(3n−1)/2   1, 5, 12, 22, 35, ...
1040.    Hexagonal    H[n]=n(2n−1)     1, 6, 15, 28, 45, ...
1041.
1042.    It can be verified that T = P = H = 40755.
1043.
1044.    Find the next triangle number that is also pentagonal and hexagonal.
1045.
1046.
1048.
1049.
1050. Problem 46
1051. ==========
1052.
1053.
1054.    It was proposed by Christian Goldbach that every odd composite number can
1055.    be written as the sum of a prime and twice a square.
1056.
1057.    9 = 7 + 2×1^2
1058.    15 = 7 + 2×2^2
1059.    21 = 3 + 2×3^2
1060.    25 = 7 + 2×3^2
1061.    27 = 19 + 2×2^2
1062.    33 = 31 + 2×1^2
1063.
1064.    It turns out that the conjecture was false.
1065.
1066.    What is the smallest odd composite that cannot be written as the sum of a
1067.    prime and twice a square?
1068.
1069.
1071.
1072.
1073. Problem 47
1074. ==========
1075.
1076.
1077.    The first two consecutive numbers to have two distinct prime factors are:
1078.
1079.    14 = 2 × 7
1080.    15 = 3 × 5
1081.
1082.    The first three consecutive numbers to have three distinct prime factors
1083.    are:
1084.
1085.    644 = 2² × 7 × 23
1086.    645 = 3 × 5 × 43
1087.    646 = 2 × 17 × 19.
1088.
1089.    Find the first four consecutive integers to have four distinct prime
1090.    factors. What is the first of these numbers?
1091.
1092.
1094.
1095.
1096. Problem 48
1097. ==========
1098.
1099.
1100.    The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
1101.
1102.    Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
1103.
1104.
1106.
1107.
1108. Problem 49
1109. ==========
1110.
1111.
1112.    The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
1113.    increases by 3330, is unusual in two ways: (i) each of the three terms are
1114.    prime, and, (ii) each of the 4-digit numbers are permutations of one
1115.    another.
1116.
1117.    There are no arithmetic sequences made up of three 1-, 2-, or 3-digit
1118.    primes, exhibiting this property, but there is one other 4-digit
1119.    increasing sequence.
1120.
1121.    What 12-digit number do you form by concatenating the three terms in this
1122.    sequence?
1123.
1124.
1126.
1127.
1128. Problem 50
1129. ==========
1130.
1131.
1132.    The prime 41, can be written as the sum of six consecutive primes:
1133.
1134.                           41 = 2 + 3 + 5 + 7 + 11 + 13
1135.
1136.    This is the longest sum of consecutive primes that adds to a prime below
1137.    one-hundred.
1138.
1139.    The longest sum of consecutive primes below one-thousand that adds to a
1140.    prime, contains 21 terms, and is equal to 953.
1141.
1142.    Which prime, below one-million, can be written as the sum of the most
1143.    consecutive primes?
1144.
1145.
1147.
1148.
1149. Problem 51
1150. ==========
1151.
1152.
1153.    By replacing the 1^st digit of the 2-digit number *3, it turns out that
1154.    six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all
1155.    prime.
1156.
1157.    By replacing the 3^rd and 4^th digits of 56**3 with the same digit, this
1158.    5-digit number is the first example having seven primes among the ten
1159.    generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663,
1160.    56773, and 56993. Consequently 56003, being the first member of this
1161.    family, is the smallest prime with this property.
1162.
1163.    Find the smallest prime which, by replacing part of the number (not
1164.    necessarily adjacent digits) with the same digit, is part of an eight
1165.    prime value family.
1166.
1167.
1169.
1170.
1171. Problem 52
1172. ==========
1173.
1174.
1175.    It can be seen that the number, 125874, and its double, 251748, contain
1176.    exactly the same digits, but in a different order.
1177.
1178.    Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x,
1179.    contain the same digits.
1180.
1181.
1183.
1184.
1185. Problem 53
1186. ==========
1187.
1188.
1189.    There are exactly ten ways of selecting three from five, 12345:
1190.
1191.               123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
1192.
1193.    In combinatorics, we use the notation, ^5C = 10.
1194.
1195.    In general,
1196.
1197.       ^nC[r] =    n!    ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
1198.                r!(n−r)!
1199.
1200.    It is not until n = 23, that a value exceeds one-million: ^23C =
1201.    1144066.
1202.
1203.    How many, not necessarily distinct, values of  ^nC[r], for 1 ≤ n ≤ 100,
1204.    are greater than one-million?
1205.
1206.
1208.
1209.
1210. Problem 54
1211. ==========
1212.
1213.
1214.    In the card game poker, a hand consists of five cards and are ranked, from
1215.    lowest to highest, in the following way:
1216.
1217.      • High Card: Highest value card.
1218.      • One Pair: Two cards of the same value.
1219.      • Two Pairs: Two different pairs.
1220.      • Three of a Kind: Three cards of the same value.
1221.      • Straight: All cards are consecutive values.
1222.      • Flush: All cards of the same suit.
1223.      • Full House: Three of a kind and a pair.
1224.      • Four of a Kind: Four cards of the same value.
1225.      • Straight Flush: All cards are consecutive values of same suit.
1226.      • Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
1227.
1228.    The cards are valued in the order:
1229.    2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
1230.
1231.    If two players have the same ranked hands then the rank made up of the
1232.    highest value wins; for example, a pair of eights beats a pair of fives
1233.    (see example 1 below). But if two ranks tie, for example, both players
1234.    have a pair of queens, then highest cards in each hand are compared (see
1235.    example 4 below); if the highest cards tie then the next highest cards are
1236.    compared, and so on.
1237.
1238.    Consider the following five hands dealt to two players:
1239.
1240.            Hand   Player 1            Player 2              Winner
1241.            1      5H 5C 6S 7S KD      2C 3S 8S 8D TD        Player 2
1242.                   Pair of Fives       Pair of Eights
1243.            2      5D 8C 9S JS AC      2C 5C 7D 8S QH        Player 1
1244.                   Highest card Ace    Highest card Queen
1245.            3      2D 9C AS AH AC      3D 6D 7D TD QD        Player 2
1246.                   Three Aces          Flush with Diamonds
1247.                   4D 6S 9H QH QC      3D 6D 7H QD QS
1248.            4      Pair of Queens      Pair of Queens        Player 1
1249.                   Highest card Nine   Highest card Seven
1250.                   2H 2D 4C 4D 4S      3C 3D 3S 9S 9D
1251.            5      Full House          Full House            Player 1
1252.                   With Three Fours    with Three Threes
1253.
1254.    The file, poker.txt, contains one-thousand random hands dealt to two
1255.    players. Each line of the file contains ten cards (separated by a single
1256.    space): the first five are Player 1's cards and the last five are Player
1257.    2's cards. You can assume that all hands are valid (no invalid characters
1258.    or repeated cards), each player's hand is in no specific order, and in
1259.    each hand there is a clear winner.
1260.
1261.    How many hands does Player 1 win?
1262.
1263.
1265.    1. poker.txt
1267.
1268.
1269. Problem 55
1270. ==========
1271.
1272.
1273.    If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
1274.
1275.    Not all numbers produce palindromes so quickly. For example,
1276.
1277.    349 + 943 = 1292,
1278.    1292 + 2921 = 4213
1279.    4213 + 3124 = 7337
1280.
1281.    That is, 349 took three iterations to arrive at a palindrome.
1282.
1283.    Although no one has proved it yet, it is thought that some numbers, like
1284.    196, never produce a palindrome. A number that never forms a palindrome
1285.    through the reverse and add process is called a Lychrel number. Due to the
1286.    theoretical nature of these numbers, and for the purpose of this problem,
1287.    we shall assume that a number is Lychrel until proven otherwise. In
1288.    addition you are given that for every number below ten-thousand, it will
1289.    either (i) become a palindrome in less than fifty iterations, or, (ii) no
1290.    one, with all the computing power that exists, has managed so far to map
1291.    it to a palindrome. In fact, 10677 is the first number to be shown to
1292.    require over fifty iterations before producing a palindrome:
1293.    4668731596684224866951378664 (53 iterations, 28-digits).
1294.
1295.    Surprisingly, there are palindromic numbers that are themselves Lychrel
1296.    numbers; the first example is 4994.
1297.
1298.    How many Lychrel numbers are there below ten-thousand?
1299.
1300.    NOTE: Wording was modified slightly on 24 April 2007 to emphasise the
1301.    theoretical nature of Lychrel numbers.
1302.
1303.
1305.
1306.
1307. Problem 56
1308. ==========
1309.
1310.
1311.    A googol (10^100) is a massive number: one followed by one-hundred zeros;
1312.    100^100 is almost unimaginably large: one followed by two-hundred zeros.
1313.    Despite their size, the sum of the digits in each number is only 1.
1314.
1315.    Considering natural numbers of the form, a^b, where a, b < 100, what is
1316.    the maximum digital sum?
1317.
1318.
1320.
1321.
1322. Problem 57
1323. ==========
1324.
1325.
1326.    It is possible to show that the square root of two can be expressed as an
1327.    infinite continued fraction.
1328.
1329.               √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
1330.
1331.    By expanding this for the first four iterations, we get:
1332.
1333.    1 + 1/2 = 3/2 = 1.5
1334.    1 + 1/(2 + 1/2) = 7/5 = 1.4
1335.    1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1336.    1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
1337.
1338.    The next three expansions are 99/70, 239/169, and 577/408, but the eighth
1339.    expansion, 1393/985, is the first example where the number of digits in
1340.    the numerator exceeds the number of digits in the denominator.
1341.
1342.    In the first one-thousand expansions, how many fractions contain a
1343.    numerator with more digits than denominator?
1344.
1345.
1347.
1348.
1349. Problem 58
1350. ==========
1351.
1352.
1353.    Starting with 1 and spiralling anticlockwise in the following way, a
1354.    square spiral with side length 7 is formed.
1355.
1356.                               37 36 35 34 33 32 31
1357.                               38 17 16 15 14 13 30
1358.                               39 18  5  4  3 12 29
1359.                               40 19  6  1  2 11 28
1360.                               41 20  7  8  9 10 27
1361.                               42 21 22 23 24 25 26
1362.                               43 44 45 46 47 48 49
1363.
1364.    It is interesting to note that the odd squares lie along the bottom right
1365.    diagonal, but what is more interesting is that 8 out of the 13 numbers
1366.    lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
1367.
1368.    If one complete new layer is wrapped around the spiral above, a square
1369.    spiral with side length 9 will be formed. If this process is continued,
1370.    what is the side length of the square spiral for which the ratio of primes
1371.    along both diagonals first falls below 10%?
1372.
1373.
1375.
1376.
1377. Problem 59
1378. ==========
1379.
1380.
1381.    Each character on a computer is assigned a unique code and the preferred
1382.    standard is ASCII (American Standard Code for Information Interchange).
1383.    For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
1384.
1385.    A modern encryption method is to take a text file, convert the bytes to
1386.    ASCII, then XOR each byte with a given value, taken from a secret key. The
1387.    advantage with the XOR function is that using the same encryption key on
1388.    the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
1389.    then 107 XOR 42 = 65.
1390.
1391.    For unbreakable encryption, the key is the same length as the plain text
1392.    message, and the key is made up of random bytes. The user would keep the
1393.    encrypted message and the encryption key in different locations, and
1394.    without both "halves", it is impossible to decrypt the message.
1395.
1396.    Unfortunately, this method is impractical for most users, so the modified
1397.    method is to use a password as a key. If the password is shorter than the
1398.    message, which is likely, the key is repeated cyclically throughout the
1399.    message. The balance for this method is using a sufficiently long password
1400.    key for security, but short enough to be memorable.
1401.
1403.    lower case characters. Using cipher1.txt, a file containing the
1404.    encrypted ASCII codes, and the knowledge that the plain text must contain
1405.    common English words, decrypt the message and find the sum of the ASCII
1406.    values in the original text.
1407.
1408.
1410.    1. cipher1.txt
1412.
1413.
1414. Problem 60
1415. ==========
1416.
1417.
1418.    The primes 3, 7, 109, and 673, are quite remarkable. By taking any two
1419.    primes and concatenating them in any order the result will always be
1420.    prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The
1421.    sum of these four primes, 792, represents the lowest sum for a set of four
1422.    primes with this property.
1423.
1424.    Find the lowest sum for a set of five primes for which any two primes
1425.    concatenate to produce another prime.
1426.
1427.
1429.
1430.
1431. Problem 61
1432. ==========
1433.
1434.
1435.    Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers
1436.    are all figurate (polygonal) numbers and are generated by the following
1437.    formulae:
1438.
1439.    Triangle     P[3,n]=n(n+1)/2    1, 3, 6, 10, 15, ...
1440.    Square       P[4,n]=n^2         1, 4, 9, 16, 25, ...
1441.    Pentagonal   P[5,n]=n(3n−1)/2   1, 5, 12, 22, 35, ...
1442.    Hexagonal    P[6,n]=n(2n−1)     1, 6, 15, 28, 45, ...
1443.    Heptagonal   P[7,n]=n(5n−3)/2   1, 7, 18, 34, 55, ...
1444.    Octagonal    P[8,n]=n(3n−2)     1, 8, 21, 40, 65, ...
1445.
1446.    The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three
1447.    interesting properties.
1448.
1449.     1. The set is cyclic, in that the last two digits of each number is the
1450.        first two digits of the next number (including the last number with
1451.        the first).
1452.     2. Each polygonal type: triangle (P[3,127]=8128), square (P[4,91]=8281),
1453.        and pentagonal (P[5,44]=2882), is represented by a different number in
1454.        the set.
1455.     3. This is the only set of 4-digit numbers with this property.
1456.
1457.    Find the sum of the only ordered set of six cyclic 4-digit numbers for
1458.    which each polygonal type: triangle, square, pentagonal, hexagonal,
1459.    heptagonal, and octagonal, is represented by a different number in the
1460.    set.
1461.
1462.
1464.
1465.
1466. Problem 62
1467. ==========
1468.
1469.
1470.    The cube, 41063625 (345^3), can be permuted to produce two other cubes:
1471.    56623104 (384^3) and 66430125 (405^3). In fact, 41063625 is the smallest
1472.    cube which has exactly three permutations of its digits which are also
1473.    cube.
1474.
1475.    Find the smallest cube for which exactly five permutations of its digits
1476.    are cube.
1477.
1478.
1480.
1481.
1482. Problem 63
1483. ==========
1484.
1485.
1486.    The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the
1487.    9-digit number, 134217728=8^9, is a ninth power.
1488.
1489.    How many n-digit positive integers exist which are also an nth power?
1490.
1491.
1493.
1494.
1495. Problem 64
1496. ==========
1497.
1498.
1499.    All square roots are periodic when written as continued fractions and can
1500.    be written in the form:
1501.
1502.    √N = a +            1
1503.                a +         1
1504.                       a +     1
1505.                              a + ...
1506.
1507.    For example, let us consider √23:
1508.
1509.    √23 = 4 + √23 — 4 = 4 +    1    = 4 +       1
1510.                               1           1 +  √23 – 3
1511.                             √23—4                 7
1512.
1513.    If we continue we would get the following expansion:
1514.
1515.    √23 = 4 +          1
1516.              1 +        1
1517.                  3 +      1
1518.                      1 +    1
1519.                          8 + ...
1520.
1521.    The process can be summarised as follows:
1522.
1523.    a = 4,     1    =   √23+4    = 1 +  √23—3
1524.                √23—4        7               7
1525.    a = 1,     7    =  7(√23+3)  = 3 +  √23—3
1526.                √23—3        14              2
1527.    a = 3,     2    =  2(√23+3)  = 1 +  √23—4
1528.                √23—3        14              7
1529.    a = 1,     7    =  7(√23+4)  = 8 +  √23—4
1530.                √23—4        7
1531.    a = 8,     1    =   √23+4    = 1 +  √23—3
1532.                √23—4        7               7
1533.    a = 1,     7    =  7(√23+3)  = 3 +  √23—3
1534.                √23—3        14              2
1535.    a = 3,     2    =  2(√23+3)  = 1 +  √23—4
1536.                √23—3        14              7
1537.    a = 1,     7    =  7(√23+4)  = 8 +  √23—4
1538.                √23—4        7
1539.
1540.    It can be seen that the sequence is repeating. For conciseness, we use the
1541.    notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats
1542.    indefinitely.
1543.
1544.    The first ten continued fraction representations of (irrational) square
1545.    roots are:
1546.
1547.    √2=[1;(2)], period=1
1548.    √3=[1;(1,2)], period=2
1549.    √5=[2;(4)], period=1
1550.    √6=[2;(2,4)], period=2
1551.    √7=[2;(1,1,1,4)], period=4
1552.    √8=[2;(1,4)], period=2
1553.    √10=[3;(6)], period=1
1554.    √11=[3;(3,6)], period=2
1555.    √12= [3;(2,6)], period=2
1556.    √13=[3;(1,1,1,1,6)], period=5
1557.
1558.    Exactly four continued fractions, for N ≤ 13, have an odd period.
1559.
1560.    How many continued fractions for N ≤ 10000 have an odd period?
1561.
1562.
1564.
1565.
1566. Problem 65
1567. ==========
1568.
1569.
1570.    The square root of 2 can be written as an infinite continued fraction.
1571.
1572.    √2 = 1 +          1
1573.             2 +        1
1574.                 2 +      1
1575.                     2 +    1
1576.                         2 + ...
1577.
1578.    The infinite continued fraction can be written, √2 = [1;(2)], (2)
1579.    indicates that 2 repeats ad infinitum. In a similar way, √23 =
1580.    [4;(1,3,1,8)].
1581.
1582.    It turns out that the sequence of partial values of continued fractions
1583.    for square roots provide the best rational approximations. Let us consider
1584.    the convergents for √2.
1585.
1586.    1 + 1 = 3/2
1587.        2
1588.
1589.    1 +   1   = 7/5
1590.        2 + 1
1591.            2
1592.
1593.    1 +     1     = 17/12
1594.        2 +   1
1595.            2 + 1
1596.                2
1597.
1598.    1 +       1       = 41/29
1599.        2 +     1
1600.            2 +   1
1601.                2 + 1
1602.                    2
1603.
1604.    Hence the sequence of the first ten convergents for √2 are:
1605.
1606.    1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378,
1607.    ...
1608.
1609.    What is most surprising is that the important mathematical constant,
1610.    e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
1611.
1612.    The first ten terms in the sequence of convergents for e are:
1613.
1614.    2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
1615.
1616.    The sum of digits in the numerator of the 10^th convergent is 1+4+5+7=17.
1617.
1618.    Find the sum of digits in the numerator of the 100^th convergent of the
1619.    continued fraction for e.
1620.
1621.
1623.
1624.
1625. Problem 66
1626. ==========
1627.
1628.
1629.    Consider quadratic Diophantine equations of the form:
1630.
1631.                                  x^2 – Dy^2 = 1
1632.
1633.    For example, when D=13, the minimal solution in x is 649^2 – 13×180^2 = 1.
1634.
1635.    It can be assumed that there are no solutions in positive integers when D
1636.    is square.
1637.
1638.    By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the
1639.    following:
1640.
1641.    3^2 – 2×2^2 = 1
1642.    2^2 – 3×1^2 = 1
1643.    9^2 – 5×4^2 = 1
1644.    5^2 – 6×2^2 = 1
1645.    8^2 – 7×3^2 = 1
1646.
1647.    Hence, by considering minimal solutions in x for D ≤ 7, the largest x is
1648.    obtained when D=5.
1649.
1650.    Find the value of D ≤ 1000 in minimal solutions of x for which the largest
1651.    value of x is obtained.
1652.
1653.
1655.
1656.
1657. Problem 67
1658. ==========
1659.
1660.
1661.    By starting at the top of the triangle below and moving to adjacent
1662.    numbers on the row below, the maximum total from top to bottom is 23.
1663.
1664.                                        3
1665.                                       7 4
1666.                                      2 4 6
1667.                                     8 5 9 3
1668.
1669.    That is, 3 + 7 + 4 + 9 = 23.
1670.
1671.    Find the maximum total from top to bottom in triangle.txt, a 15K text
1672.    file containing a triangle with one-hundred rows.
1673.
1674.    NOTE: This is a much more difficult version of Problem 18. It is not
1675.    possible to try every route to solve this problem, as there are 2^99
1676.    altogether! If you could check one trillion (10^12) routes every second it
1677.    would take over twenty billion years to check them all. There is an
1678.    efficient algorithm to solve it. ;o)
1679.
1680.
1682.    1. triangle.txt
1683.    2. problem=18
1685.
1686.
1687. Problem 68
1688. ==========
1689.
1690.
1691.    Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6,
1692.    and each line adding to nine.
1693.
1694.    Working clockwise, and starting from the group of three with the
1695.    numerically lowest external node (4,3,2 in this example), each solution
1696.    can be described uniquely. For example, the above solution can be
1697.    described by the set: 4,3,2; 6,2,1; 5,1,3.
1698.
1699.    It is possible to complete the ring with four different totals: 9, 10, 11,
1700.    and 12. There are eight solutions in total.
1701.
1702.            Total          Solution Set
1703.            9              4,2,3; 5,3,1; 6,1,2
1704.            9              4,3,2; 6,2,1; 5,1,3
1705.            10             2,3,5; 4,5,1; 6,1,3
1706.            10             2,5,3; 6,3,1; 4,1,5
1707.            11             1,4,6; 3,6,2; 5,2,4
1708.            11             1,6,4; 5,4,2; 3,2,6
1709.            12             1,5,6; 2,6,4; 3,4,5
1710.            12             1,6,5; 3,5,4; 2,4,6
1711.
1712.    By concatenating each group it is possible to form 9-digit strings; the
1713.    maximum string for a 3-gon ring is 432621513.
1714.
1715.    Using the numbers 1 to 10, and depending on arrangements, it is possible
1716.    to form 16- and 17-digit strings. What is the maximum 16-digit string for
1717.    a "magic" 5-gon ring?
1718.
1719.
1720.    p_068_1.gif
1721.    p_068_2.gif
1723.
1724.
1725. Problem 69
1726. ==========
1727.
1728.
1729.    Euler's Totient function, φ(n) [sometimes called the phi function], is
1730.    used to determine the number of numbers less than n which are relatively
1731.    prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine
1732.    and relatively prime to nine, φ(9)=6.
1733.
1734.    ┌────┬──────────────────┬──────┬───────────┐
1735.    │ n  │ Relatively Prime │ φ(n) │ n/φ(n)    │
1736.    ├────┼──────────────────┼──────┼───────────┤
1737.    │ 2  │ 1                │ 1    │ 2         │
1738.    ├────┼──────────────────┼──────┼───────────┤
1739.    │ 3  │ 1,2              │ 2    │ 1.5       │
1740.    ├────┼──────────────────┼──────┼───────────┤
1741.    │ 4  │ 1,3              │ 2    │ 2         │
1742.    ├────┼──────────────────┼──────┼───────────┤
1743.    │ 5  │ 1,2,3,4          │ 4    │ 1.25      │
1744.    ├────┼──────────────────┼──────┼───────────┤
1745.    │ 6  │ 1,5              │ 2    │ 3         │
1746.    ├────┼──────────────────┼──────┼───────────┤
1747.    │ 7  │ 1,2,3,4,5,6      │ 6    │ 1.1666... │
1748.    ├────┼──────────────────┼──────┼───────────┤
1749.    │ 8  │ 1,3,5,7          │ 4    │ 2         │
1750.    ├────┼──────────────────┼──────┼───────────┤
1751.    │ 9  │ 1,2,4,5,7,8      │ 6    │ 1.5       │
1752.    ├────┼──────────────────┼──────┼───────────┤
1753.    │ 10 │ 1,3,7,9          │ 4    │ 2.5       │
1754.    └────┴──────────────────┴──────┴───────────┘
1755.
1756.    It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
1757.
1758.    Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
1759.
1760.
1762.
1763.
1764. Problem 70
1765. ==========
1766.
1767.
1768.    Euler's Totient function, φ(n) [sometimes called the phi function], is
1769.    used to determine the number of positive numbers less than or equal to n
1770.    which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are
1771.    all less than nine and relatively prime to nine, φ(9)=6.
1772.    The number 1 is considered to be relatively prime to every positive
1773.    number, so φ(1)=1.
1774.
1775.    Interestingly, φ(87109)=79180, and it can be seen that 87109 is a
1776.    permutation of 79180.
1777.
1778.    Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n
1779.    and the ratio n/φ(n) produces a minimum.
1780.
1781.
1783.
1784.
1785. Problem 71
1786. ==========
1787.
1788.
1789.    Consider the fraction, n/d, where n and d are positive integers. If n<d
1790.    and HCF(n,d)=1, it is called a reduced proper fraction.
1791.
1792.    If we list the set of reduced proper fractions for d ≤ 8 in ascending
1793.    order of size, we get:
1794.
1795.    1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
1796.                           5/7, 3/4, 4/5, 5/6, 6/7, 7/8
1797.
1798.    It can be seen that 2/5 is the fraction immediately to the left of 3/7.
1799.
1800.    By listing the set of reduced proper fractions for d ≤ 1,000,000 in
1801.    ascending order of size, find the numerator of the fraction immediately to
1802.    the left of 3/7.
1803.
1804.
1806.
1807.
1808. Problem 72
1809. ==========
1810.
1811.
1812.    Consider the fraction, n/d, where n and d are positive integers. If n<d
1813.    and HCF(n,d)=1, it is called a reduced proper fraction.
1814.
1815.    If we list the set of reduced proper fractions for d ≤ 8 in ascending
1816.    order of size, we get:
1817.
1818.    1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
1819.                           5/7, 3/4, 4/5, 5/6, 6/7, 7/8
1820.
1821.    It can be seen that there are 21 elements in this set.
1822.
1823.    How many elements would be contained in the set of reduced proper
1824.    fractions for d ≤ 1,000,000?
1825.
1826.
1828.
1829.
1830. Problem 73
1831. ==========
1832.
1833.
1834.    Consider the fraction, n/d, where n and d are positive integers. If n<d
1835.    and HCF(n,d)=1, it is called a reduced proper fraction.
1836.
1837.    If we list the set of reduced proper fractions for d ≤ 8 in ascending
1838.    order of size, we get:
1839.
1840.    1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
1841.                           5/7, 3/4, 4/5, 5/6, 6/7, 7/8
1842.
1843.    It can be seen that there are 3 fractions between 1/3 and 1/2.
1844.
1845.    How many fractions lie between 1/3 and 1/2 in the sorted set of reduced
1846.    proper fractions for d ≤ 12,000?
1847.
1848.
1850.
1851.
1852. Problem 74
1853. ==========
1854.
1855.
1856.    The number 145 is well known for the property that the sum of the
1857.    factorial of its digits is equal to 145:
1858.
1859.    1! + 4! + 5! = 1 + 24 + 120 = 145
1860.
1861.    Perhaps less well known is 169, in that it produces the longest chain of
1862.    numbers that link back to 169; it turns out that there are only three such
1863.    loops that exist:
1864.
1865.    169 → 363601 → 1454 → 169
1866.    871 → 45361 → 871
1867.    872 → 45362 → 872
1868.
1869.    It is not difficult to prove that EVERY starting number will eventually
1870.    get stuck in a loop. For example,
1871.
1872.    69 → 363600 → 1454 → 169 → 363601 (→ 1454)
1873.    78 → 45360 → 871 → 45361 (→ 871)
1874.    540 → 145 (→ 145)
1875.
1876.    Starting with 69 produces a chain of five non-repeating terms, but the
1877.    longest non-repeating chain with a starting number below one million is
1878.    sixty terms.
1879.
1880.    How many chains, with a starting number below one million, contain exactly
1881.    sixty non-repeating terms?
1882.
1883.
1885.
1886.
1887. Problem 75
1888. ==========
1889.
1890.
1891.    It turns out that 12 cm is the smallest length of wire that can be bent to
1892.    form an integer sided right angle triangle in exactly one way, but there
1893.    are many more examples.
1894.
1895.    12 cm: (3,4,5)
1896.    24 cm: (6,8,10)
1897.    30 cm: (5,12,13)
1898.    36 cm: (9,12,15)
1899.    40 cm: (8,15,17)
1900.    48 cm: (12,16,20)
1901.
1902.    In contrast, some lengths of wire, like 20 cm, cannot be bent to form an
1903.    integer sided right angle triangle, and other lengths allow more than one
1904.    solution to be found; for example, using 120 cm it is possible to form
1905.    exactly three different integer sided right angle triangles.
1906.
1907.    120 cm: (30,40,50), (20,48,52), (24,45,51)
1908.
1909.    Given that L is the length of the wire, for how many values of L ≤
1910.    1,500,000 can exactly one integer sided right angle triangle be formed?
1911.
1912.
1914.
1915.
1916. Problem 76
1917. ==========
1918.
1919.
1920.    It is possible to write five as a sum in exactly six different ways:
1921.
1922.    4 + 1
1923.    3 + 2
1924.    3 + 1 + 1
1925.    2 + 2 + 1
1926.    2 + 1 + 1 + 1
1927.    1 + 1 + 1 + 1 + 1
1928.
1929.    How many different ways can one hundred be written as a sum of at least
1930.    two positive integers?
1931.
1932.
1934.
1935.
1936. Problem 77
1937. ==========
1938.
1939.
1940.    It is possible to write ten as the sum of primes in exactly five different
1941.    ways:
1942.
1943.    7 + 3
1944.    5 + 5
1945.    5 + 3 + 2
1946.    3 + 3 + 2 + 2
1947.    2 + 2 + 2 + 2 + 2
1948.
1949.    What is the first value which can be written as the sum of primes in over
1950.    five thousand different ways?
1951.
1952.
1954.
1955.
1956. Problem 78
1957. ==========
1958.
1959.
1960.    Let p(n) represent the number of different ways in which n coins can be
1961.    separated into piles. For example, five coins can separated into piles in
1962.    exactly seven different ways, so p(5)=7.
1963.
1964.                                OOOOO
1965.
1966.                                OOOO   O
1967.
1968.                                OOO   OO
1969.
1970.                                OOO   O   O
1971.
1972.                                OO   OO   O
1973.
1974.                                OO   O   O   O
1975.
1976.                                O   O   O   O   O
1977.
1978.    Find the least value of n for which p(n) is divisible by one million.
1979.
1980.
1982.
1983.
1984. Problem 79
1985. ==========
1986.
1987.
1988.    A common security method used for online banking is to ask the user for
1989.    three random characters from a passcode. For example, if the passcode was
1990.    531278, they may ask for the 2nd, 3rd, and 5th characters; the expected
1992.
1993.    The text file, keylog.txt, contains fifty successful login attempts.
1994.
1995.    Given that the three characters are always asked for in order, analyse the
1996.    file so as to determine the shortest possible secret passcode of unknown
1997.    length.
1998.
1999.
2001.    1. keylog.txt
2003.
2004.
2005. Problem 80
2006. ==========
2007.
2008.
2009.    It is well known that if the square root of a natural number is not an
2010.    integer, then it is irrational. The decimal expansion of such square roots
2011.    is infinite without any repeating pattern at all.
2012.
2013.    The square root of two is 1.41421356237309504880..., and the digital sum
2014.    of the first one hundred decimal digits is 475.
2015.
2016.    For the first one hundred natural numbers, find the total of the digital
2017.    sums of the first one hundred decimal digits for all the irrational square
2018.    roots.
2019.
2020.
2022.
2023.
2024. Problem 81
2025. ==========
2026.
2027.
2028.    In the 5 by 5 matrix below, the minimal path sum from the top left to the
2029.    bottom right, by only moving to the right and down, is indicated in bold
2030.    red and is equal to 2427.
2031.
2032.                               131 673 234 103 18
2033.                               201 96  342 965 150
2034.                               630 803 746 422 111
2035.                               537 699 497 121 956
2036.                               805 732 524 37  331
2037.
2038.    Find the minimal path sum, in matrix.txt, a 31K text file containing a
2039.    80 by 80 matrix, from the top left to the bottom right by only moving
2040.    right and down.
2041.
2042.
2044.    1. matrix.txt
2046.
2047.
2048. Problem 82
2049. ==========
2050.
2051.
2052.    NOTE: This problem is a more challenging version of Problem 81.
2053.
2054.    The minimal path sum in the 5 by 5 matrix below, by starting in any cell
2055.    in the left column and finishing in any cell in the right column, and only
2056.    moving up, down, and right, is indicated in red and bold; the sum is equal
2057.    to 994.
2058.
2059.                               131 673 234 103 18
2060.                               201 96  342 965 150
2061.                               630 803 746 422 111
2062.                               537 699 497 121 956
2063.                               805 732 524 37  331
2064.
2065.    Find the minimal path sum, in matrix.txt, a 31K text file containing a
2066.    80 by 80 matrix, from the left column to the right column.
2067.
2068.
2070.    1. problem=81
2071.    2. matrix.txt
2073.
2074.
2075. Problem 83
2076. ==========
2077.
2078.
2079.    NOTE: This problem is a significantly more challenging version of
2080.    Problem 81.
2081.
2082.    In the 5 by 5 matrix below, the minimal path sum from the top left to the
2083.    bottom right, by moving left, right, up, and down, is indicated in bold
2084.    red and is equal to 2297.
2085.
2086.                               131 673 234 103 18
2087.                               201 96  342 965 150
2088.                               630 803 746 422 111
2089.                               537 699 497 121 956
2090.                               805 732 524 37  331
2091.
2092.    Find the minimal path sum, in matrix.txt, a 31K text file containing a
2093.    80 by 80 matrix, from the top left to the bottom right by moving left,
2094.    right, up, and down.
2095.
2096.
2098.    1. problem=81
2099.    2. matrix.txt
2101.
2102.
2103. Problem 84
2104. ==========
2105.
2106.
2107.    In the game, Monopoly, the standard board is set up in the following way:
2108.
2109.                 GO   A1  CC1  A2  T1  R1  B1  CH1  B2   B3  JAIL
2110.                 H2                                          C1
2111.                 T2                                          U1
2112.                 H1                                          C2
2113.                 CH3                                         C3
2114.                 R4                                          R2
2115.                 G3                                          D1
2116.                 CC3                                         CC2
2117.                 G2                                          D2
2118.                 G1                                          D3
2119.                 G2J  F3  U2   F2  F1  R3  E3  E2   CH2  E1  FP
2120.
2121.    A player starts on the GO square and adds the scores on two 6-sided dice
2122.    to determine the number of squares they advance in a clockwise direction.
2123.    Without any further rules we would expect to visit each square with equal
2124.    probability: 2.5%. However, landing on G2J (Go To Jail), CC (community
2125.    chest), and CH (chance) changes this distribution.
2126.
2127.    In addition to G2J, and one card from each of CC and CH, that orders the
2128.    player to go directly to jail, if a player rolls three consecutive
2129.    doubles, they do not advance the result of their 3rd roll. Instead they
2130.    proceed directly to jail.
2131.
2132.    At the beginning of the game, the CC and CH cards are shuffled. When a
2133.    player lands on CC or CH they take a card from the top of the respective
2134.    pile and, after following the instructions, it is returned to the bottom
2135.    of the pile. There are sixteen cards in each pile, but for the purpose of
2136.    this problem we are only concerned with cards that order a movement; any
2137.    instruction not concerned with movement will be ignored and the player
2138.    will remain on the CC/CH square.
2139.
2140.      • Community Chest (2/16 cards):
2141.
2143.          2. Go to JAIL
2144.
2145.      • Chance (10/16 cards):
2146.
2148.          2. Go to JAIL
2149.          3. Go to C1
2150.          4. Go to E3
2151.          5. Go to H2
2152.          6. Go to R1
2153.          7. Go to next R (railway company)
2154.          8. Go to next R
2155.          9. Go to next U (utility company)
2156.         10. Go back 3 squares.
2157.
2158.    The heart of this problem concerns the likelihood of visiting a particular
2159.    square. That is, the probability of finishing at that square after a roll.
2160.    For this reason it should be clear that, with the exception of G2J for
2161.    which the probability of finishing on it is zero, the CH squares will have
2162.    the lowest probabilities, as 5/8 request a movement to another square, and
2163.    it is the final square that the player finishes at on each roll that we
2164.    are interested in. We shall make no distinction between "Just Visiting"
2165.    and being sent to JAIL, and we shall also ignore the rule about requiring
2166.    a double to "get out of jail", assuming that they pay to get out on their
2167.    next turn.
2168.
2169.    By starting at GO and numbering the squares sequentially from 00 to 39 we
2170.    can concatenate these two-digit numbers to produce strings that correspond
2171.    with sets of squares.
2172.
2173.    Statistically it can be shown that the three most popular squares, in
2174.    order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO
2175.    (3.09%) = Square 00. So these three most popular squares can be listed
2176.    with the six-digit modal string: 102400.
2177.
2178.    If, instead of using two 6-sided dice, two 4-sided dice are used, find the
2179.    six-digit modal string.
2180.
2181.
2183.
2184.
2185. Problem 85
2186. ==========
2187.
2188.
2189.    By counting carefully it can be seen that a rectangular grid measuring 3
2190.    by 2 contains eighteen rectangles:
2191.
2192.    Although there exists no rectangular grid that contains exactly two
2193.    million rectangles, find the area of the grid with the nearest solution.
2194.
2195.
2196.    p_085.gif
2198.
2199.
2200. Problem 86
2201. ==========
2202.
2203.
2204.    A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3,
2205.    and a fly, F, sits in the opposite corner. By travelling on the surfaces
2206.    of the room the shortest "straight line" distance from S to F is 10 and
2207.    the path is shown on the diagram.
2208.
2209.    However, there are up to three "shortest" path candidates for any given
2210.    cuboid and the shortest route doesn't always have integer length.
2211.
2212.    By considering all cuboid rooms with integer dimensions, up to a maximum
2213.    size of M by M by M, there are exactly 2060 cuboids for which the shortest
2214.    route has integer length when M=100, and this is the least value of M for
2215.    which the number of solutions first exceeds two thousand; the number of
2216.    solutions is 1975 when M=99.
2217.
2218.    Find the least value of M such that the number of solutions first exceeds
2219.    one million.
2220.
2221.
2222.    p_086.gif
2224.
2225.
2226. Problem 87
2227. ==========
2228.
2229.
2230.    The smallest number expressible as the sum of a prime square, prime cube,
2231.    and prime fourth power is 28. In fact, there are exactly four numbers
2232.    below fifty that can be expressed in such a way:
2233.
2234.    28 = 2^2 + 2^3 + 2^4
2235.    33 = 3^2 + 2^3 + 2^4
2236.    49 = 5^2 + 2^3 + 2^4
2237.    47 = 2^2 + 3^3 + 2^4
2238.
2239.    How many numbers below fifty million can be expressed as the sum of a
2240.    prime square, prime cube, and prime fourth power?
2241.
2242.
2244.
2245.
2246. Problem 88
2247. ==========
2248.
2249.
2250.    A natural number, N, that can be written as the sum and product of a given
2251.    set of at least two natural numbers, {a, a, ... , a[k]} is called a
2252.    product-sum number: N = a + a + ... + a[k] = a × a × ... ×
2253.    a[k].
2254.
2255.    For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
2256.
2257.    For a given set of size, k, we shall call the smallest N with this
2258.    property a minimal product-sum number. The minimal product-sum numbers for
2259.    sets of size, k = 2, 3, 4, 5, and 6 are as follows.
2260.
2261.    k=2: 4 = 2 × 2 = 2 + 2
2262.    k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
2263.    k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
2264.    k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
2265.    k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
2266.
2267.    Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is
2268.    4+6+8+12 = 30; note that 8 is only counted once in the sum.
2269.
2270.    In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is
2271.    {4, 6, 8, 12, 15, 16}, the sum is 61.
2272.
2273.    What is the sum of all the minimal product-sum numbers for 2≤k≤12000?
2274.
2275.
2277.
2278.
2279. Problem 89
2280. ==========
2281.
2282.
2283.    The rules for writing Roman numerals allow for many ways of writing each
2284.    number (see About Roman Numerals...). However, there is always a "best"
2285.    way of writing a particular number.
2286.
2287.    For example, the following represent all of the legitimate ways of writing
2288.    the number sixteen:
2289.
2290.    IIIIIIIIIIIIIIII
2291.    VIIIIIIIIIII
2292.    VVIIIIII
2293.    XIIIIII
2294.    VVVI
2295.    XVI
2296.
2297.    The last example being considered the most efficient, as it uses the least
2298.    number of numerals.
2299.
2300.    The 11K text file, roman.txt, contains one thousand numbers written in
2301.    valid, but not necessarily minimal, Roman numerals; that is, they are
2302.    arranged in descending units and obey the subtractive pair rule (see
2303.    About Roman Numerals... for the definitive rules for this problem).
2304.
2305.    Find the number of characters saved by writing each of these in their
2306.    minimal form.
2307.
2308.    Note: You can assume that all the Roman numerals in the file contain no
2309.    more than four consecutive identical units.
2310.
2311.
2314.    2. roman.txt
2317.
2318.
2319. Problem 90
2320. ==========
2321.
2322.
2323.    Each of the six faces on a cube has a different digit (0 to 9) written on
2324.    it; the same is done to a second cube. By placing the two cubes
2325.    side-by-side in different positions we can form a variety of 2-digit
2326.    numbers.
2327.
2328.    For example, the square number 64 could be formed:
2329.
2330.    In fact, by carefully choosing the digits on both cubes it is possible to
2331.    display all of the square numbers below one-hundred: 01, 04, 09, 16, 25,
2332.    36, 49, 64, and 81.
2333.
2334.    For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9}
2335.    on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
2336.
2337.    However, for this problem we shall allow the 6 or 9 to be turned
2338.    upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3,
2339.    4, 6, 7} allows for all nine square numbers to be displayed; otherwise it
2340.    would be impossible to obtain 09.
2341.
2342.    In determining a distinct arrangement we are interested in the digits on
2343.    each cube, not the order.
2344.
2345.    {1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
2346.    {1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
2347.
2348.    But because we are allowing 6 and 9 to be reversed, the two distinct sets
2349.    in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9}
2350.    for the purpose of forming 2-digit numbers.
2351.
2352.    How many distinct arrangements of the two cubes allow for all of the
2353.    square numbers to be displayed?
2354.
2355.
2356.    p_090.gif
2358.
2359.
2360. Problem 91
2361. ==========
2362.
2363.
2364.    The points P (x, y) and Q (x, y) are plotted at integer
2365.    co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.
2366.
2367.    There are exactly fourteen triangles containing a right angle that can be
2368.    formed when each co-ordinate lies between 0 and 2 inclusive; that is,
2369.    0 ≤ x, y, x, y ≤ 2.
2370.
2371.    Given that 0 ≤ x, y, x, y ≤ 50, how many right triangles can
2372.    be formed?
2373.
2374.
2375.    p_091_1.gif
2376.    p_091_2.gif
2378.
2379.
2380. Problem 92
2381. ==========
2382.
2383.
2384.    A number chain is created by continuously adding the square of the digits
2385.    in a number to form a new number until it has been seen before.
2386.
2387.    For example,
2388.
2389.    44 → 32 → 13 → 10 → 1 → 1
2390.    85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89
2391.
2392.    Therefore any chain that arrives at 1 or 89 will become stuck in an
2393.    endless loop. What is most amazing is that EVERY starting number will
2394.    eventually arrive at 1 or 89.
2395.
2396.    How many starting numbers below ten million will arrive at 89?
2397.
2398.
2400.
2401.
2402. Problem 93
2403. ==========
2404.
2405.
2406.    By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and
2407.    making use of the four arithmetic operations (+, −, *, /) and
2408.    brackets/parentheses, it is possible to form different positive integer
2409.    targets.
2410.
2411.    For example,
2412.
2413.    8 = (4 * (1 + 3)) / 2
2414.    14 = 4 * (3 + 1 / 2)
2415.    19 = 4 * (2 + 3) − 1
2416.    36 = 3 * 4 * (2 + 1)
2417.
2418.    Note that concatenations of the digits, like 12 + 34, are not allowed.
2419.
2420.    Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different
2421.    target numbers of which 36 is the maximum, and each of the numbers 1 to 28
2422.    can be obtained before encountering the first non-expressible number.
2423.
2424.    Find the set of four distinct digits, a < b < c < d, for which the longest
2425.    set of consecutive positive integers, 1 to n, can be obtained, giving your
2426.    answer as a string: abcd.
2427.
2428.
2430.
2431.
2432. Problem 94
2433. ==========
2434.
2435.
2436.    It is easily proved that no equilateral triangle exists with integral
2437.    length sides and integral area. However, the almost equilateral triangle
2438.    5-5-6 has an area of 12 square units.
2439.
2440.    We shall define an almost equilateral triangle to be a triangle for which
2441.    two sides are equal and the third differs by no more than one unit.
2442.
2443.    Find the sum of the perimeters of all almost equilateral triangles with
2444.    integral side lengths and area and whose perimeters do not exceed one
2445.    billion (1,000,000,000).
2446.
2447.
2449.
2450.
2451. Problem 95
2452. ==========
2453.
2454.
2455.    The proper divisors of a number are all the divisors excluding the number
2456.    itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As
2457.    the sum of these divisors is equal to 28, we call it a perfect number.
2458.
2459.    Interestingly the sum of the proper divisors of 220 is 284 and the sum of
2460.    the proper divisors of 284 is 220, forming a chain of two numbers. For
2461.    this reason, 220 and 284 are called an amicable pair.
2462.
2463.    Perhaps less well known are longer chains. For example, starting with
2464.    12496, we form a chain of five numbers:
2465.
2466.              12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)
2467.
2468.    Since this chain returns to its starting point, it is called an amicable
2469.    chain.
2470.
2471.    Find the smallest member of the longest amicable chain with no element
2472.    exceeding one million.
2473.
2474.
2476.
2477.
2478. Problem 96
2479. ==========
2480.
2481.
2482.    Su Doku (Japanese meaning number place) is the name given to a popular
2483.    puzzle concept. Its origin is unclear, but credit must be attributed to
2484.    Leonhard Euler who invented a similar, and much more difficult, puzzle
2485.    idea called Latin Squares. The objective of Su Doku puzzles, however, is
2486.    to replace the blanks (or zeros) in a 9 by 9 grid in such that each row,
2487.    column, and 3 by 3 box contains each of the digits 1 to 9. Below is an
2488.    example of a typical starting puzzle grid and its solution grid.
2489.
2490.           ┌───────┬───────┬───────┐         ┌───────┬───────┬───────┐
2491.           │ 0 0 3 │ 0 2 0 │ 6 0 0 │         │ 4 8 3 │ 9 2 1 │ 6 5 7 │
2492.           │ 9 0 0 │ 3 0 5 │ 0 0 1 │         │ 9 6 7 │ 3 4 5 │ 8 2 1 │
2493.           │ 0 0 1 │ 8 0 6 │ 4 0 0 │         │ 2 5 1 │ 8 7 6 │ 4 9 3 │
2494.           ├───────┼───────┼───────┤         ├───────┼───────┼───────┤
2495.           │ 0 0 8 │ 1 0 2 │ 9 0 0 │         │ 5 4 8 │ 1 3 2 │ 9 7 6 │
2496.           │ 7 0 0 │ 0 0 0 │ 0 0 8 │         │ 7 2 9 │ 5 6 4 │ 1 3 8 │
2497.           │ 0 0 6 │ 7 0 8 │ 2 0 0 │         │ 1 3 6 │ 7 9 8 │ 2 4 5 │
2498.           ├───────┼───────┼───────┤         ├───────┼───────┼───────┤
2499.           │ 0 0 2 │ 6 0 9 │ 5 0 0 │         │ 3 7 2 │ 6 8 9 │ 5 1 4 │
2500.           │ 8 0 0 │ 2 0 3 │ 0 0 9 │         │ 8 1 4 │ 2 5 3 │ 7 6 9 │
2501.           │ 0 0 5 │ 0 1 0 │ 3 0 0 │         │ 6 9 5 │ 4 1 7 │ 3 8 2 │
2502.           └───────┴───────┴───────┘         └───────┴───────┴───────┘
2503.
2504.    A well constructed Su Doku puzzle has a unique solution and can be solved
2505.    by logic, although it may be necessary to employ "guess and test" methods
2506.    in order to eliminate options (there is much contested opinion over this).
2507.    The complexity of the search determines the difficulty of the puzzle; the
2508.    example above is considered easy because it can be solved by straight
2509.    forward direct deduction.
2510.
2511.    The 6K text file, sudoku.txt, contains fifty different Su Doku puzzles
2512.    ranging in difficulty, but all with unique solutions (the first puzzle in
2513.    the file is the example above).
2514.
2515.    By solving all fifty puzzles find the sum of the 3-digit numbers found in
2516.    the top left corner of each solution grid; for example, 483 is the 3-digit
2517.    number found in the top left corner of the solution grid above.
2518.
2519.
2521.    1. sudoku.txt
2523.
2524.
2525. Problem 97
2526. ==========
2527.
2528.
2529.    The first known prime found to exceed one million digits was discovered in
2530.    1999, and is a Mersenne prime of the form 2^6972593−1; it contains exactly
2531.    2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p−1,
2532.    have been found which contain more digits.
2533.
2534.    However, in 2004 there was found a massive non-Mersenne prime which
2535.    contains 2,357,207 digits: 28433×2^7830457+1.
2536.
2537.    Find the last ten digits of this prime number.
2538.
2539.
2541.
2542.
2543. Problem 98
2544. ==========
2545.
2546.
2547.    By replacing each of the letters in the word CARE with 1, 2, 9, and 6
2548.    respectively, we form a square number: 1296 = 36^2. What is remarkable is
2549.    that, by using the same digital substitutions, the anagram, RACE, also
2550.    forms a square number: 9216 = 96^2. We shall call CARE (and RACE) a square
2551.    anagram word pair and specify further that leading zeroes are not
2552.    permitted, neither may a different letter have the same digital value as
2553.    another letter.
2554.
2555.    Using words.txt, a 16K text file containing nearly two-thousand common
2556.    English words, find all the square anagram word pairs (a palindromic word
2557.    is NOT considered to be an anagram of itself).
2558.
2559.    What is the largest square number formed by any member of such a pair?
2560.
2561.    NOTE: All anagrams formed must be contained in the given text file.
2562.
2563.
2565.    1. words.txt
2567.
2568.
2569. Problem 99
2570. ==========
2571.
2572.
2573.    Comparing two numbers written in index form like 2^11 and 3^7 is not
2574.    difficult, as any calculator would confirm that 2^11 = 2048 < 3^7 = 2187.
2575.
2576.    However, confirming that 632382^518061 > 519432^525806 would be much more
2577.    difficult, as both numbers contain over three million digits.
2578.
2579.    Using base_exp.txt, a 22K text file containing one thousand lines with
2580.    a base/exponent pair on each line, determine which line number has the
2581.    greatest numerical value.
2582.
2583.    NOTE: The first two lines in the file represent the numbers in the example
2584.    given above.
2585.
2586.
2588.    1. base_exp.txt
2590.
2591.
2592. Problem 100
2593. ===========
2594.
2595.
2596.    If a box contains twenty-one coloured discs, composed of fifteen blue
2597.    discs and six red discs, and two discs were taken at random, it can be
2598.    seen that the probability of taking two blue discs, P(BB) =
2599.    (15/21)×(14/20) = 1/2.
2600.
2601.    The next such arrangement, for which there is exactly 50% chance of taking
2602.    two blue discs at random, is a box containing eighty-five blue discs and
2603.    thirty-five red discs.
2604.
2605.    By finding the first arrangement to contain over 10^12 = 1,000,000,000,000
2606.    discs in total, determine the number of blue discs that the box would
2607.    contain.
2608.
2609.
2611.
2612.
2613. Problem 101
2614. ===========
2615.
2616.
2617.    If we are presented with the first k terms of a sequence it is impossible
2618.    to say with certainty the value of the next term, as there are infinitely
2619.    many polynomial functions that can model the sequence.
2620.
2621.    As an example, let us consider the sequence of cube numbers. This is
2622.    defined by the generating function,
2623.    u[n] = n^3: 1, 8, 27, 64, 125, 216, ...
2624.
2625.    Suppose we were only given the first two terms of this sequence. Working
2626.    on the principle that "simple is best" we should assume a linear
2627.    relationship and predict the next term to be 15 (common difference 7).
2628.    Even if we were presented with the first three terms, by the same
2629.    principle of simplicity, a quadratic relationship should be assumed.
2630.
2631.    We shall define OP(k, n) to be the n^th term of the optimum polynomial
2632.    generating function for the first k terms of a sequence. It should be
2633.    clear that OP(k, n) will accurately generate the terms of the sequence for
2634.    n ≤ k, and potentially the first incorrect term (FIT) will be OP(k, k+1);
2635.    in which case we shall call it a bad OP (BOP).
2636.
2637.    As a basis, if we were only given the first term of sequence, it would be
2638.    most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u.
2639.
2640.    Hence we obtain the following OPs for the cubic sequence:
2641.
2642.    OP(1, n) = 1               1, 1, 1, 1, ...
2643.    OP(2, n) = 7n−6            1, 8, 15, ...
2644.    OP(3, n) = 6n^2−11n+6      1, 8, 27, 58, ...
2645.    OP(4, n) = n^3             1, 8, 27, 64, 125, ...
2646.
2647.    Clearly no BOPs exist for k ≥ 4.
2648.
2649.    By considering the sum of FITs generated by the BOPs (indicated in red
2650.    above), we obtain 1 + 15 + 58 = 74.
2651.
2652.    Consider the following tenth degree polynomial generating function:
2653.
2654.       u[n] = 1 − n + n^2 − n^3 + n^4 − n^5 + n^6 − n^7 + n^8 − n^9 + n^10
2655.
2656.    Find the sum of FITs for the BOPs.
2657.
2658.
2660.
2661.
2662. Problem 102
2663. ===========
2664.
2665.
2666.    Three distinct points are plotted at random on a Cartesian plane, for
2667.    which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.
2668.
2669.    Consider the following two triangles:
2670.
2671.                      A(-340,495), B(-153,-910), C(835,-947)
2672.
2673.                      X(-175,41), Y(-421,-714), Z(574,-645)
2674.
2675.    It can be verified that triangle ABC contains the origin, whereas triangle
2676.    XYZ does not.
2677.
2678.    Using triangles.txt, a 27K text file containing the co-ordinates of one
2679.    thousand "random" triangles, find the number of triangles for which the
2680.    interior contains the origin.
2681.
2682.    NOTE: The first two examples in the file represent the triangles in the
2683.    example given above.
2684.
2685.
2687.    1. triangles.txt
2689.
2690.
2691. Problem 103
2692. ===========
2693.
2694.
2695.    Let S(A) represent the sum of elements in set A of size n. We shall call
2696.    it a special sum set if for any two non-empty disjoint subsets, B and C,
2697.    the following properties are true:
2698.
2699.    i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2700.    ii. If B contains more elements than C then S(B) > S(C).
2701.
2702.    If S(A) is minimised for a given n, we shall call it an optimum special
2703.    sum set. The first five optimum special sum sets are given below.
2704.
2705.    n = 1: {1}
2706.    n = 2: {1, 2}
2707.    n = 3: {2, 3, 4}
2708.    n = 4: {3, 5, 6, 7}
2709.    n = 5: {6, 9, 11, 12, 13}
2710.
2711.    It seems that for a given optimum set, A = {a, a, ... , a[n]}, the
2712.    next optimum set is of the form B = {b, a+b, a+b, ... ,a[n]+b},
2713.    where b is the "middle" element on the previous row.
2714.
2715.    By applying this "rule" we would expect the optimum set for n = 6 to be A
2716.    = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the
2717.    optimum set, as we have merely applied an algorithm to provide a near
2718.    optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25},
2719.    with S(A) = 115 and corresponding set string: 111819202225.
2720.
2721.    Given that A is an optimum special sum set for n = 7, find its set string.
2722.
2723.    NOTE: This problem is related to Problem 105 and Problem 106.
2724.
2725.
2727.    1. problem=105
2728.    2. problem=106
2730.
2731.
2732. Problem 104
2733. ===========
2734.
2735.
2736.    The Fibonacci sequence is defined by the recurrence relation:
2737.
2738.      F[n] = F[n−1] + F[n−2], where F = 1 and F = 1.
2739.
2740.    It turns out that F, which contains 113 digits, is the first
2741.    Fibonacci number for which the last nine digits are 1-9 pandigital
2742.    (contain all the digits 1 to 9, but not necessarily in order). And
2743.    F, which contains 575 digits, is the first Fibonacci number for
2744.    which the first nine digits are 1-9 pandigital.
2745.
2746.    Given that F[k] is the first Fibonacci number for which the first nine
2747.    digits AND the last nine digits are 1-9 pandigital, find k.
2748.
2749.
2751.
2752.
2753. Problem 105
2754. ===========
2755.
2756.
2757.    Let S(A) represent the sum of elements in set A of size n. We shall call
2758.    it a special sum set if for any two non-empty disjoint subsets, B and C,
2759.    the following properties are true:
2760.
2761.    i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2762.    ii. If B contains more elements than C then S(B) > S(C).
2763.
2764.    For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set
2765.    because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159,
2766.    161, 139, 158} satisfies both rules for all possible subset pair
2767.    combinations and S(A) = 1286.
2768.
2769.    Using sets.txt (right click and "Save Link/Target As..."), a 4K text
2770.    file with one-hundred sets containing seven to twelve elements (the two
2771.    examples given above are the first two sets in the file), identify all the
2772.    special sum sets, A, A, ..., A[k], and find the value of S(A) +
2773.    S(A) + ... + S(A[k]).
2774.
2775.    NOTE: This problem is related to Problem 103 and Problem 106.
2776.
2777.
2779.    1. sets.txt
2780.    2. problem=103
2781.    3. problem=106
2783.
2784.
2785. Problem 106
2786. ===========
2787.
2788.
2789.    Let S(A) represent the sum of elements in set A of size n. We shall call
2790.    it a special sum set if for any two non-empty disjoint subsets, B and C,
2791.    the following properties are true:
2792.
2793.    i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2794.    ii. If B contains more elements than C then S(B) > S(C).
2795.
2796.    For this problem we shall assume that a given set contains n strictly
2797.    increasing elements and it already satisfies the second rule.
2798.
2799.    Surprisingly, out of the 25 possible subset pairs that can be obtained
2800.    from a set for which n = 4, only 1 of these pairs need to be tested for
2801.    equality (first rule). Similarly, when n = 7, only 70 out of the 966
2802.    subset pairs need to be tested.
2803.
2804.    For n = 12, how many of the 261625 subset pairs that can be obtained need
2805.    to be tested for equality?
2806.
2807.    NOTE: This problem is related to Problem 103 and Problem 105.
2808.
2809.
2811.    1. problem=103
2812.    2. problem=105
2814.
2815.
2816. Problem 107
2817. ===========
2818.
2819.
2820.    The following undirected network consists of seven vertices and twelve
2821.    edges with a total weight of 243.
2822.
2823.    The same network can be represented by the matrix below.
2824.
2825.                   ┌──────┬────┬────┬────┬────┬────┬────┬────┐
2826.                   │      │ A  │ B  │ C  │ D  │ E  │ F  │ G  │
2827.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2828.                   │ A    │ -  │ 16 │ 12 │ 21 │ -  │ -  │ -  │
2829.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2830.                   │ B    │ 16 │ -  │ -  │ 17 │ 20 │ -  │ -  │
2831.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2832.                   │ C    │ 12 │ -  │ -  │ 28 │ -  │ 31 │ -  │
2833.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2834.                   │ D    │ 21 │ 17 │ 28 │ -  │ 18 │ 19 │ 23 │
2835.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2836.                   │ E    │ -  │ 20 │ -  │ 18 │ -  │ -  │ 11 │
2837.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2838.                   │ F    │ -  │ -  │ 31 │ 19 │ -  │ -  │ 27 │
2839.                   ├──────┼────┼────┼────┼────┼────┼────┼────┤
2840.                   │ G    │ -  │ -  │ -  │ 23 │ 11 │ 27 │ -  │
2841.                   └──────┴────┴────┴────┴────┴────┴────┴────┘
2842.
2843.    However, it is possible to optimise the network by removing some edges and
2844.    still ensure that all points on the network remain connected. The network
2845.    which achieves the maximum saving is shown below. It has a weight of 93,
2846.    representing a saving of 243 − 93 = 150 from the original network.
2847.
2848.    Using network.txt, a 6K text file containing a network with forty
2849.    vertices, and given in matrix form, find the maximum saving which can be
2850.    achieved by removing redundant edges whilst ensuring that the network
2851.    remains connected.
2852.
2853.
2855.    1. network.txt
2856.    p_107_1.gif
2857.    p_107_2.gif
2859.
2860.
2861. Problem 108
2862. ===========
2863.
2864.
2865.    In the following equation x, y, and n are positive integers.
2866.
2867.                                    1   1   1
2868.                                    ─ + ─ = ─
2869.                                    x   y   n
2870.
2871.    For n = 4 there are exactly three distinct solutions:
2872.
2873.                                    1   1    1
2874.                                    ─ + ─  = ─
2875.                                    5   20   4
2876.                                    1   1    1
2877.                                    ─ + ─  = ─
2878.                                    6   12   4
2879.                                    1   1    1
2880.                                    ─ + ─  = ─
2881.                                    8   8    4
2882.
2883.    What is the least value of n for which the number of distinct solutions
2884.    exceeds one-thousand?
2885.
2886.    NOTE: This problem is an easier version of Problem 110; it is strongly
2887.    advised that you solve this one first.
2888.
2889.
2891.    1. problem=110
2893.
2894.
2895. Problem 109
2896. ===========
2897.
2898.
2899.    In the game of darts a player throws three darts at a target board which
2900.    is split into twenty equal sized sections numbered one to twenty.
2901.
2902.    The score of a dart is determined by the number of the region that the
2903.    dart lands in. A dart landing outside the red/green outer ring scores
2904.    zero. The black and cream regions inside this ring represent single
2905.    scores. However, the red/green outer ring and middle ring score double and
2906.    treble scores respectively.
2907.
2908.    At the centre of the board are two concentric circles called the bull
2909.    region, or bulls-eye. The outer bull is worth 25 points and the inner bull
2910.    is a double, worth 50 points.
2911.
2912.    There are many variations of rules but in the most popular game the
2913.    players will begin with a score 301 or 501 and the first player to reduce
2914.    their running total to zero is a winner. However, it is normal to play a
2915.    "doubles out" system, which means that the player must land a double
2916.    (including the double bulls-eye at the centre of the board) on their final
2917.    dart to win; any other dart that would reduce their running total to one
2918.    or lower means the score for that set of three darts is "bust".
2919.
2920.    When a player is able to finish on their current score it is called a
2921.    "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s
2922.    and double bull).
2923.
2924.    There are exactly eleven distinct ways to checkout on a score of 6:
2925.
2926.                                    ┌──┬──┬──┐
2927.                                    │D3│  │  │
2928.                                    ├──┼──┼──┤
2929.                                    │D1│D2│  │
2930.                                    ├──┼──┼──┤
2931.                                    │S2│D2│  │
2932.                                    ├──┼──┼──┤
2933.                                    │D2│D1│  │
2934.                                    ├──┼──┼──┤
2935.                                    │S4│D1│  │
2936.                                    ├──┼──┼──┤
2937.                                    │S1│S1│D2│
2938.                                    ├──┼──┼──┤
2939.                                    │S1│T1│D1│
2940.                                    ├──┼──┼──┤
2941.                                    │S1│S3│D1│
2942.                                    ├──┼──┼──┤
2943.                                    │D1│D1│D1│
2944.                                    ├──┼──┼──┤
2945.                                    │D1│S2│D1│
2946.                                    ├──┼──┼──┤
2947.                                    │S2│S2│D1│
2948.                                    └──┴──┴──┘
2949.
2950.    Note that D1 D2 is considered different to D2 D1 as they finish on
2951.    different doubles. However, the combination S1 T1 D1 is considered the
2952.    same as T1 S1 D1.
2953.
2954.    In addition we shall not include misses in considering combinations; for
2955.    example, D3 is the same as 0 D3 and 0 0 D3.
2956.
2957.    Incredibly there are 42336 distinct ways of checking out in total.
2958.
2959.    How many distinct ways can a player checkout with a score less than 100?
2960.
2961.
2962.    p_109.gif
2964.
2965.
2966. Problem 110
2967. ===========
2968.
2969.
2970.    In the following equation x, y, and n are positive integers.
2971.
2972.                                    1   1   1
2973.                                    ─ + ─ = ─
2974.                                    x   y   n
2975.
2976.    It can be verified that when n = 1260 there are 113 distinct solutions and
2977.    this is the least value of n for which the total number of distinct
2978.    solutions exceeds one hundred.
2979.
2980.    What is the least value of n for which the number of distinct solutions
2981.    exceeds four million?
2982.
2983.    NOTE: This problem is a much more difficult version of Problem 108 and
2984.    as it is well beyond the limitations of a brute force approach it requires
2985.    a clever implementation.
2986.
2987.
2989.    1. problem=108
2991.
2992.
2993. Problem 111
2994. ===========
2995.
2996.
2997.    Considering 4-digit primes containing repeated digits it is clear that
2998.    they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by
2999.    22, and so on. But there are nine 4-digit primes containing three ones:
3000.
3001.               1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
3002.
3003.    We shall say that M(n, d) represents the maximum number of repeated digits
3004.    for an n-digit prime where d is the repeated digit, N(n, d) represents the
3005.    number of such primes, and S(n, d) represents the sum of these primes.
3006.
3007.    So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit
3008.    prime where one is the repeated digit, there are N(4, 1) = 9 such primes,
3009.    and the sum of these primes is S(4, 1) = 22275. It turns out that for d =
3010.    0, it is only possible to have M(4, 0) = 2 repeated digits, but there are
3011.    N(4, 0) = 13 such cases.
3012.
3013.    In the same way we obtain the following results for 4-digit primes.
3014.
3015.                    ┌──────────┬─────────┬─────────┬─────────┐
3016.                    │ Digit, d │ M(4, d) │ N(4, d) │ S(4, d) │
3017.                    ├──────────┼─────────┼─────────┼─────────┤
3018.                    │ 0        │ 2       │ 13      │ 67061   │
3019.                    ├──────────┼─────────┼─────────┼─────────┤
3020.                    │ 1        │ 3       │ 9       │ 22275   │
3021.                    ├──────────┼─────────┼─────────┼─────────┤
3022.                    │ 2        │ 3       │ 1       │ 2221    │
3023.                    ├──────────┼─────────┼─────────┼─────────┤
3024.                    │ 3        │ 3       │ 12      │ 46214   │
3025.                    ├──────────┼─────────┼─────────┼─────────┤
3026.                    │ 4        │ 3       │ 2       │ 8888    │
3027.                    ├──────────┼─────────┼─────────┼─────────┤
3028.                    │ 5        │ 3       │ 1       │ 5557    │
3029.                    ├──────────┼─────────┼─────────┼─────────┤
3030.                    │ 6        │ 3       │ 1       │ 6661    │
3031.                    ├──────────┼─────────┼─────────┼─────────┤
3032.                    │ 7        │ 3       │ 9       │ 57863   │
3033.                    ├──────────┼─────────┼─────────┼─────────┤
3034.                    │ 8        │ 3       │ 1       │ 8887    │
3035.                    ├──────────┼─────────┼─────────┼─────────┤
3036.                    │ 9        │ 3       │ 7       │ 48073   │
3037.                    └──────────┴─────────┴─────────┴─────────┘
3038.
3039.    For d = 0 to 9, the sum of all S(4, d) is 273700.
3040.
3041.    Find the sum of all S(10, d).
3042.
3043.
3045.
3046.
3047. Problem 112
3048. ===========
3049.
3050.
3051.    Working from left-to-right if no digit is exceeded by the digit to its
3052.    left it is called an increasing number; for example, 134468.
3053.
3054.    Similarly if no digit is exceeded by the digit to its right it is called a
3055.    decreasing number; for example, 66420.
3056.
3057.    We shall call a positive integer that is neither increasing nor decreasing
3058.    a "bouncy" number; for example, 155349.
3059.
3060.    Clearly there cannot be any bouncy numbers below one-hundred, but just
3061.    over half of the numbers below one-thousand (525) are bouncy. In fact, the
3062.    least number for which the proportion of bouncy numbers first reaches 50%
3063.    is 538.
3064.
3065.    Surprisingly, bouncy numbers become more and more common and by the time
3066.    we reach 21780 the proportion of bouncy numbers is equal to 90%.
3067.
3068.    Find the least number for which the proportion of bouncy numbers is
3069.    exactly 99%.
3070.
3071.
3073.
3074.
3075. Problem 113
3076. ===========
3077.
3078.
3079.    Working from left-to-right if no digit is exceeded by the digit to its
3080.    left it is called an increasing number; for example, 134468.
3081.
3082.    Similarly if no digit is exceeded by the digit to its right it is called a
3083.    decreasing number; for example, 66420.
3084.
3085.    We shall call a positive integer that is neither increasing nor decreasing
3086.    a "bouncy" number; for example, 155349.
3087.
3088.    As n increases, the proportion of bouncy numbers below n increases such
3089.    that there are only 12951 numbers below one-million that are not bouncy
3090.    and only 277032 non-bouncy numbers below 10^10.
3091.
3092.    How many numbers below a googol (10^100) are not bouncy?
3093.
3094.
3096.
3097.
3098. Problem 114
3099. ===========
3100.
3101.
3102.    A row measuring seven units in length has red blocks with a minimum length
3103.    of three units placed on it, such that any two red blocks (which are
3104.    allowed to be different lengths) are separated by at least one black
3105.    square. There are exactly seventeen ways of doing this.
3106.
3107.                           ┌┬┬┬┬┬┬┐  ┌──┬┬┬┬┐  ┌┬──┬┬┬┐
3108.                           └┴┴┴┴┴┴┘  └──┴┴┴┴┘  └┴──┴┴┴┘
3109.                           ┌┬┬──┬┬┐  ┌┬┬┬──┬┐  ┌┬┬┬┬──┐
3110.                           └┴┴──┴┴┘  └┴┴┴──┴┘  └┴┴┴┴──┘
3111.                           ┌──┬┬──┐  ┌───┬┬┬┐  ┌┬───┬┬┐
3112.                           └──┴┴──┘  └───┴┴┴┘  └┴───┴┴┘
3113.                           ┌┬┬───┬┐  ┌┬┬┬───┐  ┌────┬┬┐
3114.                           └┴┴───┴┘  └┴┴┴───┘  └────┴┴┘
3115.                           ┌┬────┬┐  ┌┬┬────┐  ┌─────┬┐
3116.                           └┴────┴┘  └┴┴────┘  └─────┴┘
3117.                           ┌┬─────┐  ┌──────┐
3118.                           └┴─────┘  └──────┘
3119.
3120.    How many ways can a row measuring fifty units in length be filled?
3121.
3122.    NOTE: Although the example above does not lend itself to the possibility,
3123.    in general it is permitted to mix block sizes. For example, on a row
3124.    measuring eight units in length you could use red (3), black (1), and red
3125.    (4).
3126.
3127.
3129.
3130.
3131. Problem 115
3132. ===========
3133.
3134.
3135.    NOTE: This is a more difficult version of Problem 114.
3136.
3137.    A row measuring n units in length has red blocks with a minimum length of
3138.    m units placed on it, such that any two red blocks (which are allowed to
3139.    be different lengths) are separated by at least one black square.
3140.
3141.    Let the fill-count function, F(m, n), represent the number of ways that a
3142.    row can be filled.
3143.
3144.    For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
3145.
3146.    That is, for m = 3, it can be seen that n = 30 is the smallest value for
3147.    which the fill-count function first exceeds one million.
3148.
3149.    In the same way, for m = 10, it can be verified that F(10, 56) = 880711
3150.    and F(10, 57) = 1148904, so n = 57 is the least value for which the
3151.    fill-count function first exceeds one million.
3152.
3153.    For m = 50, find the least value of n for which the fill-count function
3154.    first exceeds one million.
3155.
3156.
3158.    1. problem=114
3160.
3161.
3162. Problem 116
3163. ===========
3164.
3165.
3166.    A row of five black square tiles is to have a number of its tiles replaced
3167.    with coloured oblong tiles chosen from red (length two), green (length
3168.    three), or blue (length four).
3169.
3170.    If red tiles are chosen there are exactly seven ways this can be done.
3171.
3172.                          ┌─╥╥╥┐  ┌╥─╥╥┐  ┌╥╥─╥┐  ┌╥╥╥─┐
3173.                          └─╨╨╨┘  └╨─╨╨┘  └╨╨─╨┘  └╨╨╨─┘
3174.
3175.                          ┌─╥─╥┐  ┌─╥╥─┐  ┌╥─╥─┐
3176.                          └─╨─╨┘  └─╨╨─┘  └╨─╨─┘
3177.
3178.    If green tiles are chosen there are three ways.
3179.
3180.                            ┌──╥╥┐  ┌╥──╥┐  ┌╥╥──┐
3181.                            └──╨╨┘  └╨──╨┘  └╨╨──┘
3182.
3183.    And if blue tiles are chosen there are two ways.
3184.
3185.                                  ┌╥───┐  ┌───╥┐
3186.                                  └╨───┘  └───╨┘
3187.
3188.    Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of
3189.    replacing the black tiles in a row measuring five units in length.
3190.
3191.    How many different ways can the black tiles in a row measuring fifty units
3192.    in length be replaced if colours cannot be mixed and at least one coloured
3193.    tile must be used?
3194.
3195.    NOTE: This is related to Problem 117.
3196.
3197.
3199.    1. problem=117
3201.
3202.
3203. Problem 117
3204. ===========
3205.
3206.
3207.    Using a combination of black square tiles and oblong tiles chosen from:
3208.    red tiles measuring two units, green tiles measuring three units, and blue
3209.    tiles measuring four units, it is possible to tile a row measuring five
3210.    units in length in exactly fifteen different ways.
3211.
3212.                          ┌╥╥╥╥┐  ┌─╥╥╥┐  ┌╥─╥╥┐  ┌╥╥─╥┐
3213.                          └╨╨╨╨┘  └─╨╨╨┘  └╨─╨╨┘  └╨╨─╨┘
3214.
3215.                          ┌╥╥╥─┐  ┌─╥─╥┐  ┌─╥╥─┐  ┌╥─╥─┐
3216.                          └╨╨╨─┘  └─╨─╨┘  └─╨╨─┘  └╨─╨─┘
3217.
3218.                          ┌──╥╥┐  ┌╥──╥┐  ┌╥╥──┐  ┌─╥──┐
3219.                          └──╨╨┘  └╨──╨┘  └╨╨──┘  └─╨──┘
3220.
3221.                          ┌──╥─┐  ┌───╥┐  ┌╥───┐
3222.                          └──╨─┘  └───╨┘  └╨───┘
3223.
3224.    How many ways can a row measuring fifty units in length be tiled?
3225.
3226.    NOTE: This is related to Problem 116.
3227.
3228.
3230.    1. problem=116
3232.
3233.
3234. Problem 118
3235. ===========
3236.
3237.
3238.    Using all of the digits 1 through 9 and concatenating them freely to form
3239.    decimal integers, different sets can be formed. Interestingly with the set
3240.    {2,5,47,89,631}, all of the elements belonging to it are prime.
3241.
3242.    How many distinct sets containing each of the digits one through nine
3243.    exactly once contain only prime elements?
3244.
3245.
3247.
3248.
3249. Problem 119
3250. ===========
3251.
3252.
3253.    The number 512 is interesting because it is equal to the sum of its digits
3254.    raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a
3255.    number with this property is 614656 = 28^4.
3256.
3257.    We shall define a[n] to be the nth term of this sequence and insist that a
3258.    number must contain at least two digits to have a sum.
3259.
3260.    You are given that a = 512 and a = 614656.
3261.
3262.    Find a.
3263.
3264.
3266.
3267.
3268. Problem 120
3269. ===========
3270.
3271.
3272.    Let r be the remainder when (a−1)^n + (a+1)^n is divided by a^2.
3273.
3274.    For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49.
3275.    And as n varies, so too will r, but for a = 7 it turns out that r[max] =
3276.    42.
3277.
3278.    For 3 ≤ a ≤ 1000, find ∑ r[max].
3279.
3280.
3282.
3283.
3284. Problem 121
3285. ===========
3286.
3287.
3288.    A bag contains one red disc and one blue disc. In a game of chance a
3289.    player takes a disc at random and its colour is noted. After each turn the
3290.    disc is returned to the bag, an extra red disc is added, and another disc
3291.    is taken at random.
3292.
3293.    The player pays £1 to play and wins if they have taken more blue discs
3294.    than red discs at the end of the game.
3295.
3296.    If the game is played for four turns, the probability of a player winning
3297.    is exactly 11/120, and so the maximum prize fund the banker should
3298.    allocate for winning in this game would be £10 before they would expect to
3299.    incur a loss. Note that any payout will be a whole number of pounds and
3300.    also includes the original £1 paid to play the game, so in the example
3301.    given the player actually wins £9.
3302.
3303.    Find the maximum prize fund that should be allocated to a single game in
3304.    which fifteen turns are played.
3305.
3306.
3308.
3309.
3310. Problem 122
3311. ===========
3312.
3313.
3314.    The most naive way of computing n^15 requires fourteen multiplications:
3315.
3316.    n × n × ... × n = n^15
3317.
3318.    But using a "binary" method you can compute it in six multiplications:
3319.
3320.    n × n = n^2
3321.    n^2 × n^2 = n^4
3322.    n^4 × n^4 = n^8
3323.    n^8 × n^4 = n^12
3324.    n^12 × n^2 = n^14
3325.    n^14 × n = n^15
3326.
3327.    However it is yet possible to compute it in only five multiplications:
3328.
3329.    n × n = n^2
3330.    n^2 × n = n^3
3331.    n^3 × n^3 = n^6
3332.    n^6 × n^6 = n^12
3333.    n^12 × n^3 = n^15
3334.
3335.    We shall define m(k) to be the minimum number of multiplications to
3336.    compute n^k; for example m(15) = 5.
3337.
3338.    For 1 ≤ k ≤ 200, find ∑ m(k).
3339.
3340.
3342.
3343.
3344. Problem 123
3345. ===========
3346.
3347.
3348.    Let p[n] be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder
3349.    when (p[n]−1)^n + (p[n]+1)^n is divided by p[n]^2.
3350.
3351.    For example, when n = 3, p = 5, and 4^3 + 6^3 = 280 ≡ 5 mod 25.
3352.
3353.    The least value of n for which the remainder first exceeds 10^9 is 7037.
3354.
3355.    Find the least value of n for which the remainder first exceeds 10^10.
3356.
3357.
3359.
3360.
3361. Problem 124
3362. ===========
3363.
3364.
3365.    The radical of n, rad(n), is the product of the distinct prime factors of
3366.    n. For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42.
3367.
3368.    If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and
3369.    sorting on n if the radical values are equal, we get:
3370.
3371.                             Unsorted       Sorted
3373.                             1    1      1    1    1
3374.                             2    2      2    2    2
3375.                             3    3      4    2    3
3376.                             4    2      8    2    4
3377.                             5    5      3    3    5
3378.                             6    6      9    3    6
3379.                             7    7      5    5    7
3380.                             8    2      6    6    8
3381.                             9    3      7    7    9
3382.                             10   10     10   10   10
3383.
3384.    Let E(k) be the kth element in the sorted n column; for example, E(4) = 8
3385.    and E(6) = 9.
3386.
3387.    If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).
3388.
3389.
3391.
3392.
3393. Problem 125
3394. ===========
3395.
3396.
3397.    The palindromic number 595 is interesting because it can be written as the
3398.    sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.
3399.
3400.    There are exactly eleven palindromes below one-thousand that can be
3401.    written as consecutive square sums, and the sum of these palindromes is
3402.    4164. Note that 1 = 0^2 + 1^2 has not been included as this problem is
3403.    concerned with the squares of positive integers.
3404.
3405.    Find the sum of all the numbers less than 10^8 that are both palindromic
3406.    and can be written as the sum of consecutive squares.
3407.
3408.
3410.
3411.
3412. Problem 126
3413. ===========
3414.
3415.
3416.    The minimum number of cubes to cover every visible face on a cuboid
3417.    measuring 3 x 2 x 1 is twenty-two.
3418.
3419.    If we then add a second layer to this solid it would require forty-six
3420.    cubes to cover every visible face, the third layer would require
3421.    seventy-eight cubes, and the fourth layer would require one-hundred and
3422.    eighteen cubes to cover every visible face.
3423.
3424.    However, the first layer on a cuboid measuring 5 x 1 x 1 also requires
3425.    twenty-two cubes; similarly the first layer on cuboids measuring
3426.    5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.
3427.
3428.    We shall define C(n) to represent the number of cuboids that contain n
3429.    cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118)
3430.    = 8.
3431.
3432.    It turns out that 154 is the least value of n for which C(n) = 10.
3433.
3434.    Find the least value of n for which C(n) = 1000.
3435.
3436.
3437.    p_126.gif
3439.
3440.
3441. Problem 127
3442. ===========
3443.
3444.
3445.    The radical of n, rad(n), is the product of distinct prime factors of n.
3446.    For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42.
3447.
3448.    We shall define the triplet of positive integers (a, b, c) to be an
3449.    abc-hit if:
3450.
3451.     1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
3452.     2. a < b
3453.     3. a + b = c
3455.
3456.    For example, (5, 27, 32) is an abc-hit, because:
3457.
3458.     1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
3459.     2. 5 < 27
3460.     3. 5 + 27 = 32
3461.     4. rad(4320) = 30 < 32
3462.
3463.    It turns out that abc-hits are quite rare and there are only thirty-one
3464.    abc-hits for c < 1000, with ∑c = 12523.
3465.
3466.    Find ∑c for c < 120000.
3467.
3468.
3470.
3471.
3472. Problem 128
3473. ===========
3474.
3475.
3476.    A hexagonal tile with number 1 is surrounded by a ring of six hexagonal
3477.    tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an
3478.    anti-clockwise direction.
3479.
3480.    New rings are added in the same fashion, with the next rings being
3481.    numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows
3482.    the first three rings.
3483.
3484.    By finding the difference between tile n and each its six neighbours we
3485.    shall define PD(n) to be the number of those differences which are prime.
3486.
3487.    For example, working clockwise around tile 8 the differences are 12, 29,
3488.    11, 6, 1, and 13. So PD(8) = 3.
3489.
3490.    In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and
3491.    10, hence PD(17) = 2.
3492.
3493.    It can be shown that the maximum value of PD(n) is 3.
3494.
3495.    If all of the tiles for which PD(n) = 3 are listed in ascending order to
3496.    form a sequence, the 10th tile would be 271.
3497.
3498.    Find the 2000th tile in this sequence.
3499.
3500.
3501.    p_128.gif
3503.
3504.
3505. Problem 129
3506. ===========
3507.
3508.
3509.    A number consisting entirely of ones is called a repunit. We shall define
3510.    R(k) to be a repunit of length k; for example, R(6) = 111111.
3511.
3512.    Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
3513.    that there always exists a value, k, for which R(k) is divisible by n, and
3514.    let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
3515.    5.
3516.
3517.    The least value of n for which A(n) first exceeds ten is 17.
3518.
3519.    Find the least value of n for which A(n) first exceeds one-million.
3520.
3521.
3523.
3524.
3525. Problem 130
3526. ===========
3527.
3528.
3529.    A number consisting entirely of ones is called a repunit. We shall define
3530.    R(k) to be a repunit of length k; for example, R(6) = 111111.
3531.
3532.    Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
3533.    that there always exists a value, k, for which R(k) is divisible by n, and
3534.    let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
3535.    5.
3536.
3537.    You are given that for all primes, p > 5, that p − 1 is divisible by A(p).
3538.    For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
3539.
3540.    However, there are rare composite values for which this is also true; the
3541.    first five examples being 91, 259, 451, 481, and 703.
3542.
3543.    Find the sum of the first twenty-five composite values of n for which
3544.    GCD(n, 10) = 1 and n − 1 is divisible by A(n).
3545.
3546.
3548.
3549.
3550. Problem 131
3551. ===========
3552.
3553.
3554.    There are some prime values, p, for which there exists a positive integer,
3555.    n, such that the expression n^3 + n^2p is a perfect cube.
3556.
3557.    For example, when p = 19, 8^3 + 8^2×19 = 12^3.
3558.
3559.    What is perhaps most surprising is that for each prime with this property
3560.    the value of n is unique, and there are only four such primes below
3561.    one-hundred.
3562.
3563.    How many primes below one million have this remarkable property?
3564.
3565.
3567.
3568.
3569. Problem 132
3570. ===========
3571.
3572.
3573.    A number consisting entirely of ones is called a repunit. We shall define
3574.    R(k) to be a repunit of length k.
3575.
3576.    For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these
3577.    prime factors is 9414.
3578.
3579.    Find the sum of the first forty prime factors of R(10^9).
3580.
3581.
3583.
3584.
3585. Problem 133
3586. ===========
3587.
3588.
3589.    A number consisting entirely of ones is called a repunit. We shall define
3590.    R(k) to be a repunit of length k; for example, R(6) = 111111.
3591.
3592.    Let us consider repunits of the form R(10^n).
3593.
3594.    Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is
3595.    divisible by 17. Yet there is no value of n for which R(10^n) will divide
3596.    by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four
3597.    primes below one-hundred that can be a factor of R(10^n).
3598.
3599.    Find the sum of all the primes below one-hundred thousand that will never
3600.    be a factor of R(10^n).
3601.
3602.
3604.
3605.
3606. Problem 134
3607. ===========
3608.
3609.
3610.    Consider the consecutive primes p = 19 and p = 23. It can be
3611.    verified that 1219 is the smallest number such that the last digits are
3612.    formed by p whilst also being divisible by p.
3613.
3614.    In fact, with the exception of p = 3 and p = 5, for every pair of
3615.    consecutive primes, p > p, there exist values of n for which the
3616.    last digits are formed by p and n is divisible by p. Let S be the
3617.    smallest of these values of n.
3618.
3619.    Find ∑ S for every pair of consecutive primes with 5 ≤ p ≤ 1000000.
3620.
3621.
3623.
3624.
3625. Problem 135
3626. ===========
3627.
3628.
3629.    Given the positive integers, x, y, and z, are consecutive terms of an
3630.    arithmetic progression, the least value of the positive integer, n, for
3631.    which the equation, x^2 − y^2 − z^2 = n, has exactly two solutions is n =
3632.    27:
3633.
3634.                    34^2 − 27^2 − 20^2 = 12^2 − 9^2 − 6^2 = 27
3635.
3636.    It turns out that n = 1155 is the least value which has exactly ten
3637.    solutions.
3638.
3639.    How many values of n less than one million have exactly ten distinct
3640.    solutions?
3641.
3642.
3644.
3645.
3646. Problem 136
3647. ===========
3648.
3649.
3650.    The positive integers, x, y, and z, are consecutive terms of an arithmetic
3651.    progression. Given that n is a positive integer, the equation, x^2 − y^2 −
3652.    z^2 = n, has exactly one solution when n = 20:
3653.
3654.                              13^2 − 10^2 − 7^2 = 20
3655.
3656.    In fact there are twenty-five values of n below one hundred for which the
3657.    equation has a unique solution.
3658.
3659.    How many values of n less than fifty million have exactly one solution?
3660.
3661.
3663.
3664.
3665. Problem 137
3666. ===========
3667.
3668.
3669.    Consider the infinite polynomial series A[F](x) = xF + x^2F +
3670.    x^3F + ..., where F[k] is the kth term in the Fibonacci sequence: 1, 1,
3671.    2, 3, 5, 8, ... ; that is, F[k] = F[k−1] + F[k−2], F = 1 and F = 1.
3672.
3673.    For this problem we shall be interested in values of x for which A[F](x)
3674.    is a positive integer.
3675.
3676.    Surprisingly A[F](1/2)  =  (1/2).1 + (1/2)^2.1 + (1/2)^3.2 + (1/2)^4.3 +
3677.                               (1/2)^5.5 + ...
3678.                            =  1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
3679.                            =  2
3680.
3681.    The corresponding values of x for the first five natural numbers are shown
3682.    below.
3683.
3684.                               ┌─────────┬───────┐
3685.                               │x        │A[F](x)│
3686.                               ├─────────┼───────┤
3687.                               │√2−1     │1      │
3688.                               ├─────────┼───────┤
3689.                               │1/2      │2      │
3690.                               ├─────────┼───────┤
3691.                               │(√13−2)/3│3      │
3692.                               ├─────────┼───────┤
3693.                               │(√89−5)/8│4      │
3694.                               ├─────────┼───────┤
3695.                               │(√34−3)/5│5      │
3696.                               └─────────┴───────┘
3697.
3698.    We shall call A[F](x) a golden nugget if x is rational, because they
3699.    become increasingly rarer; for example, the 10th golden nugget is
3700.    74049690.
3701.
3702.    Find the 15th golden nugget.
3703.
3704.
3706.
3707.
3708. Problem 138
3709. ===========
3710.
3711.
3712.    Consider the isosceles triangle with base length, b = 16, and legs, L =
3713.    17.
3714.
3715.    By using the Pythagorean theorem it can be seen that the height of the
3716.    triangle, h = √(17^2 − 8^2) = 15, which is one less than the base length.
3717.
3718.    With b = 272 and L = 305, we get h = 273, which is one more than the base
3719.    length, and this is the second smallest isosceles triangle with the
3720.    property that h = b ± 1.
3721.
3722.    Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1
3723.    and b, L are positive integers.
3724.
3725.
3726.    p_138.gif
3728.
3729.
3730. Problem 139
3731. ===========
3732.
3733.
3734.    Let (a, b, c) represent the three sides of a right angle triangle with
3735.    integral length sides. It is possible to place four such triangles
3736.    together to form a square with length c.
3737.
3738.    For example, (3, 4, 5) triangles can be placed together to form a 5 by 5
3739.    square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5
3740.    square can be tiled with twenty-five 1 by 1 squares.
3741.
3742.    However, if (5, 12, 13) triangles were used then the hole would measure 7
3743.    by 7 and these could not be used to tile the 13 by 13 square.
3744.
3745.    Given that the perimeter of the right triangle is less than one-hundred
3746.    million, how many Pythagorean triangles would allow such a tiling to take
3747.    place?
3748.
3749.
3750.    p_139.gif
3752.
3753.
3754. Problem 140
3755. ===========
3756.
3757.
3758.    Consider the infinite polynomial series A[G](x) = xG + x^2G +
3759.    x^3G + ..., where G[k] is the kth term of the second order recurrence
3760.    relation G[k] = G[k−1] + G[k−2], G = 1 and G = 4; that is, 1, 4, 5,
3761.    9, 14, 23, ... .
3762.
3763.    For this problem we shall be concerned with values of x for which A[G](x)
3764.    is a positive integer.
3765.
3766.    The corresponding values of x for the first five natural numbers are shown
3767.    below.
3768.
3769.                              ┌───────────┬───────┐
3770.                              │x          │A[G](x)│
3771.                              ├───────────┼───────┤
3772.                              │(√5−1)/4   │1      │
3773.                              ├───────────┼───────┤
3774.                              │2/5        │2      │
3775.                              ├───────────┼───────┤
3776.                              │(√22−2)/6  │3      │
3777.                              ├───────────┼───────┤
3778.                              │(√137−5)/14│4      │
3779.                              ├───────────┼───────┤
3780.                              │1/2        │5      │
3781.                              └───────────┴───────┘
3782.
3783.    We shall call A[G](x) a golden nugget if x is rational, because they
3784.    become increasingly rarer; for example, the 20th golden nugget is
3785.    211345365.
3786.
3787.    Find the sum of the first thirty golden nuggets.
3788.
3789.
3791.
3792.
3793. Problem 141
3794. ===========
3795.
3796.
3797.    A positive integer, n, is divided by d and the quotient and remainder are
3798.    q and r respectively. In addition d, q, and r are consecutive positive
3799.    integer terms in a geometric sequence, but not necessarily in that order.
3800.
3801.    For example, 58 divided by 6 has quotient 9 and remainder 4. It can also
3802.    be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common
3803.    ratio 3/2).
3804.    We will call such numbers, n, progressive.
3805.
3806.    Some progressive numbers, such as 9 and 10404 = 102^2, happen to also be
3807.    perfect squares.
3808.    The sum of all progressive perfect squares below one hundred thousand is
3809.    124657.
3810.
3811.    Find the sum of all progressive perfect squares below one trillion
3812.    (10^12).
3813.
3814.
3816.
3817.
3818. Problem 142
3819. ===========
3820.
3821.
3822.    Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x
3823.    − y, x + z, x − z, y + z, y − z are all perfect squares.
3824.
3825.
3827.
3828.
3829. Problem 143
3830. ===========
3831.
3832.
3833.    Let ABC be a triangle with all interior angles being less than 120
3834.    degrees. Let X be any point inside the triangle and let XA = p, XC = q,
3835.    and XB = r.
3836.
3837.    Fermat challenged Torricelli to find the position of X such that p + q + r
3838.    was minimised.
3839.
3840.    Torricelli was able to prove that if equilateral triangles AOB, BNC and
3841.    AMC are constructed on each side of triangle ABC, the circumscribed
3842.    circles of AOB, BNC, and AMC will intersect at a single point, T, inside
3843.    the triangle. Moreover he proved that T, called the Torricelli/Fermat
3844.    point, minimises p + q + r. Even more remarkable, it can be shown that
3845.    when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO
3846.    also intersect at T.
3847.
3848.    If the sum is minimised and a, b, c, p, q and r are all positive integers
3849.    we shall call triangle ABC a Torricelli triangle. For example, a = 399, b
3850.    = 455, c = 511 is an example of a Torricelli triangle, with p + q + r =
3851.    784.
3852.
3853.    Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli
3854.    triangles.
3855.
3856.
3857.    p_143_torricelli.gif
3859.
3860.
3861. Problem 144
3862. ===========
3863.
3864.
3865.    In laser physics, a "white cell" is a mirror system that acts as a delay
3866.    line for the laser beam. The beam enters the cell, bounces around on the
3867.    mirrors, and eventually works its way back out.
3868.
3869.    The specific white cell we will be considering is an ellipse with the
3870.    equation 4x^2 + y^2 = 100
3871.
3872.    The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing,
3873.    allowing the light to enter and exit through the hole.
3874.
3875.    The light beam in this problem starts at the point (0.0,10.1) just outside
3876.    the white cell, and the beam first impacts the mirror at (1.4,-9.6).
3877.
3878.    Each time the laser beam hits the surface of the ellipse, it follows the
3879.    usual law of reflection "angle of incidence equals angle of reflection."
3880.    That is, both the incident and reflected beams make the same angle with
3881.    the normal line at the point of incidence.
3882.
3883.    In the figure on the left, the red line shows the first two points of
3884.    contact between the laser beam and the wall of the white cell; the blue
3885.    line shows the line tangent to the ellipse at the point of incidence of
3886.    the first bounce.
3887.
3888.    The slope m of the tangent line at any point (x,y) of the given ellipse
3889.    is: m = −4x/y
3890.
3891.    The normal line is perpendicular to this tangent line at the point of
3892.    incidence.
3893.
3894.    The animation on the right shows the first 10 reflections of the beam.
3895.
3896.    How many times does the beam hit the internal surface of the white cell
3897.    before exiting?
3898.
3899.
3900.    p_144_1.gif
3901.    p_144_2.gif
3903.
3904.
3905. Problem 145
3906. ===========
3907.
3908.
3909.    Some positive integers n have the property that the sum [ n + reverse(n) ]
3910.    consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and
3911.    409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409,
3912.    and 904 are reversible. Leading zeroes are not allowed in either n or
3913.    reverse(n).
3914.
3915.    There are 120 reversible numbers below one-thousand.
3916.
3917.    How many reversible numbers are there below one-billion (10^9)?
3918.
3919.
3921.
3922.
3923. Problem 146
3924. ===========
3925.
3926.
3927.    The smallest positive integer n for which the numbers n^2+1, n^2+3, n^2+7,
3928.    n^2+9, n^2+13, and n^2+27 are consecutive primes is 10. The sum of all
3929.    such integers n below one-million is 1242490.
3930.
3931.    What is the sum of all such integers n below 150 million?
3932.
3933.
3935.
3936.
3937. Problem 147
3938. ===========
3939.
3940.
3941.    In a 3x2 cross-hatched grid, a total of 37 different rectangles could be
3942.    situated within that grid as indicated in the sketch.
3943.
3944.    There are 5 grids smaller than 3x2, vertical and horizontal dimensions
3945.    being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is
3946.    cross-hatched, the following number of different rectangles could be
3947.    situated within those smaller grids:
3948.
3949.    1x1: 1
3950.    2x1: 4
3951.    3x1: 8
3952.    1x2: 4
3953.    2x2: 18
3954.
3955.    Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles
3956.    could be situated within 3x2 and smaller grids.
3957.
3958.    How many different rectangles could be situated within 47x43 and smaller
3959.    grids?
3960.
3961.
3962.    p_147.gif
3964.
3965.
3966. Problem 148
3967. ===========
3968.
3969.
3970.    We can easily verify that none of the entries in the first seven rows of
3971.    Pascal's triangle are divisible by 7:
3972.
3973.                                        1
3974.                                     1     1
3975.                                  1     2     1
3976.                               1     3     3     1
3977.                            1     4     6     4     1
3978.                         1     5    10    10     5     1
3979.                      1     6    15    20    15     6     1
3980.
3981.    However, if we check the first one hundred rows, we will find that only
3982.    2361 of the 5050 entries are not divisible by 7.
3983.
3984.    Find the number of entries which are not divisible by 7 in the first one
3985.    billion (10^9) rows of Pascal's triangle.
3986.
3987.
3989.
3990.
3991. Problem 149
3992. ===========
3993.
3994.
3995.    Looking at the table below, it is easy to verify that the maximum possible
3996.    sum of adjacent numbers in any direction (horizontal, vertical, diagonal
3997.    or anti-diagonal) is 16 (= 8 + 7 + 1).
3998.
3999.                              ┌────┬────┬────┬─────┐
4000.                              │ −2 │ 5  │ 3  │ 2   │
4001.                              ├────┼────┼────┼─────┤
4002.                              │ 9  │ −6 │ 5  │ 1   │
4003.                              ├────┼────┼────┼─────┤
4004.                              │ 3  │ 2  │ 7  │ 3   │
4005.                              ├────┼────┼────┼─────┤
4006.                              │ −1 │ 8  │ −4 │   8 │
4007.                              └────┴────┴────┴─────┘
4008.
4009.    Now, let us repeat the search, but on a much larger scale:
4010.
4011.    First, generate four million pseudo-random numbers using a specific form
4012.    of what is known as a "Lagged Fibonacci Generator":
4013.
4014.    For 1 ≤ k ≤ 55, s[k] = [100003 − 200003k + 300007k^3] (modulo 1000000) −
4015.    500000.
4016.    For 56 ≤ k ≤ 4000000, s[k] = [s[k−24] + s[k−55] + 1000000] (modulo
4017.    1000000) − 500000.
4018.
4019.    Thus, s = −393027 and s = 86613.
4020.
4021.    The terms of s are then arranged in a 2000×2000 table, using the first
4022.    2000 numbers to fill the first row (sequentially), the next 2000 numbers
4023.    to fill the second row, and so on.
4024.
4025.    Finally, find the greatest sum of (any number of) adjacent entries in any
4026.    direction (horizontal, vertical, diagonal or anti-diagonal).
4027.
4028.
4030.
4031.
4032. Problem 150
4033. ===========
4034.
4035.
4036.    In a triangular array of positive and negative integers, we wish to find a
4037.    sub-triangle such that the sum of the numbers it contains is the smallest
4038.    possible.
4039.
4040.    In the example below, it can be easily verified that the marked triangle
4041.    satisfies this condition having a sum of −42.
4042.
4043.    We wish to make such a triangular array with one thousand rows, so we
4044.    generate 500500 pseudo-random numbers s[k] in the range ±2^19, using a
4045.    type of random number generator (known as a Linear Congruential Generator)
4046.    as follows:
4047.
4048.    t := 0
4049.    for k = 1 up to k = 500500:
4050.        t := (615949*t + 797807) modulo 2^20
4051.        s[k] := t−2^19
4052.
4053.    Thus: s = 273519, s = −153582, s = 450905 etc
4054.
4055.    Our triangular array is then formed using the pseudo-random numbers thus:
4056.
4057.                                       s
4058.                                    s  s
4059.                                s  s  s
4060.                             s  s  s  s
4061.                                       ...
4062.
4063.    Sub-triangles can start at any element of the array and extend down as far
4064.    as we like (taking-in the two elements directly below it from the next
4065.    row, the three elements directly below from the row after that, and so
4066.    on).
4067.    The "sum of a sub-triangle" is defined as the sum of all the elements it
4068.    contains.
4069.    Find the smallest possible sub-triangle sum.
4070.
4071.
4072.    p_150.gif
4074.
4075.
4076. Problem 151
4077. ===========
4078.
4079.
4080.    A printing shop runs 16 batches (jobs) every week and each batch requires
4081.    a sheet of special colour-proofing paper of size A5.
4082.
4083.    Every Monday morning, the foreman opens a new envelope, containing a large
4084.    sheet of the special paper with size A1.
4085.
4086.    He proceeds to cut it in half, thus getting two sheets of size A2. Then he
4087.    cuts one of them in half to get two sheets of size A3 and so on until he
4088.    obtains the A5-size sheet needed for the first batch of the week.
4089.
4090.    All the unused sheets are placed back in the envelope.
4091.
4092.    At the beginning of each subsequent batch, he takes from the envelope one
4093.    sheet of paper at random. If it is of size A5, he uses it. If it is
4094.    larger, he repeats the 'cut-in-half' procedure until he has what he needs
4095.    and any remaining sheets are always placed back in the envelope.
4096.
4097.    Excluding the first and last batch of the week, find the expected number
4098.    of times (during each week) that the foreman finds a single sheet of paper
4099.    in the envelope.
4100.
4101.    Give your answer rounded to six decimal places using the format x.xxxxxx .
4102.
4103.
4104.    p_151.gif
4106.
4107.
4108. Problem 152
4109. ===========
4110.
4111.
4112.    There are several ways to write the number 1/2 as a sum of inverse squares
4113.    using distinct integers.
4114.
4115.    For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:
4116.
4117.    In fact, only using integers between 2 and 45 inclusive, there are exactly
4118.    three ways to do it, the remaining two being:
4119.    {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}.
4120.
4121.    How many ways are there to write the number 1/2 as a sum of inverse
4122.    squares using distinct integers between 2 and 80 inclusive?
4123.
4124.
4125.    p_152_sum.gif
4127.
4128.
4129. Problem 153
4130. ===========
4131.
4132.
4133.    As we all know the equation x^2=-1 has no solutions for real x.
4134.    If we however introduce the imaginary number i this equation has two
4135.    solutions: x=i and x=-i.
4136.    If we go a step further the equation (x-3)^2=-4 has two complex solutions:
4137.    x=3+2i and x=3-2i.
4138.    x=3+2i and x=3-2i are called each others' complex conjugate.
4139.    Numbers of the form a+bi are called complex numbers.
4140.    In general a+bi and a−bi are each other's complex conjugate.
4141.
4142.    A Gaussian Integer is a complex number a+bi such that both a and b are
4143.    integers.
4144.    The regular integers are also Gaussian integers (with b=0).
4145.    To distinguish them from Gaussian integers with b ≠ 0 we call such
4146.    integers "rational integers."
4147.    A Gaussian integer is called a divisor of a rational integer n if the
4148.    result is also a Gaussian integer.
4149.    If for example we divide 5 by 1+2i we can simplify in the following
4150.    manner:
4151.    Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
4152.    The result is .
4153.    So 1+2i is a divisor of 5.
4154.    Note that 1+i is not a divisor of 5 because .
4155.    Note also that if the Gaussian Integer (a+bi) is a divisor of a rational
4156.    integer n, then its complex conjugate (a−bi) is also a divisor of n.
4157.
4158.    In fact, 5 has six divisors such that the real part is positive: {1, 1 +
4159.    2i, 1 − 2i, 2 + i, 2 − i, 5}.
4160.    The following is a table of all of the divisors for the first five
4161.    positive rational integers:
4162.
4163.               ┌───┬──────────────────────────────┬───────────────┐
4164.               │ n │ Gaussian integer divisors    │ Sum s(n) of   │
4165.               │   │ with positive real part      │ thesedivisors │
4166.               ├───┼──────────────────────────────┼───────────────┤
4167.               │ 1 │ 1                            │ 1             │
4168.               ├───┼──────────────────────────────┼───────────────┤
4169.               │ 2 │ 1, 1+i, 1-i, 2               │ 5             │
4170.               ├───┼──────────────────────────────┼───────────────┤
4171.               │ 3 │ 1, 3                         │ 4             │
4172.               ├───┼──────────────────────────────┼───────────────┤
4173.               │ 4 │ 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 │ 13            │
4174.               ├───┼──────────────────────────────┼───────────────┤
4175.               │ 5 │ 1, 1+2i, 1-2i, 2+i, 2-i, 5   │ 12            │
4176.               └───┴──────────────────────────────┴───────────────┘
4177.
4178.    For divisors with positive real parts, then, we have: .
4179.
4180.    For 1 ≤ n ≤ 10^5, ∑ s(n)=17924657155.
4181.
4182.    What is ∑ s(n) for 1 ≤ n ≤ 10^8?
4183.
4184.
4185.    p_153_formule1.gif
4186.    p_153_formule2.gif
4187.    p_153_formule5.gif
4188.    p_153_formule6.gif
4190.
4191.
4192. Problem 154
4193. ===========
4194.
4195.
4196.    A triangular pyramid is constructed using spherical balls so that each
4197.    ball rests on exactly three balls of the next lower level.
4198.
4199.    Then, we calculate the number of paths leading from the apex to each
4200.    position:
4201.
4202.    A path starts at the apex and progresses downwards to any of the three
4203.    spheres directly below the current position.
4204.
4205.    Consequently, the number of paths to reach a certain position is the sum
4206.    of the numbers immediately above it (depending on the position, there are
4207.    up to three numbers above it).
4208.
4209.    The result is Pascal's pyramid and the numbers at each level n are the
4210.    coefficients of the trinomial expansion (x + y + z)^n.
4211.
4212.    How many coefficients in the expansion of (x + y + z)^200000 are multiples
4213.    of 10^12?
4214.
4215.
4216.    p_154_pyramid.gif
4218.
4219.
4220. Problem 155
4221. ===========
4222.
4223.
4224.    An electric circuit uses exclusively identical capacitors of the same
4225.    value C.
4226.    The capacitors can be connected in series or in parallel to form
4227.    sub-units, which can then be connected in series or in parallel with other
4228.    capacitors or other sub-units to form larger sub-units, and so on up to a
4229.    final circuit.
4230.
4231.    Using this simple procedure and up to n identical capacitors, we can make
4232.    circuits having a range of different total capacitances. For example,
4233.    using up to n=3 capacitors of 60 F each, we can obtain the following 7
4234.    distinct total capacitance values:
4235.
4236.    If we denote by D(n) the number of distinct total capacitance values we
4237.    can obtain when using up to n equal-valued capacitors and the simple
4238.    procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ...
4239.
4240.    Find D(18).
4241.
4242.    Reminder : When connecting capacitors C, C etc in parallel, the
4243.    total capacitance is C[T] = C + C +...,
4244.    whereas when connecting them in series, the overall capacitance is given
4245.    by:
4246.
4247.
4248.    p_155_capsmu.gif
4249.    p_155_capacitors1.gif
4250.    p_155_capsform.gif
4252.
4253.
4254. Problem 156
4255. ===========
4256.
4257.
4258.    Starting from zero the natural numbers are written down in base 10 like
4259.    this:
4260.    0 1 2 3 4 5 6 7 8 9 10 11 12....
4261.
4262.    Consider the digit d=1. After we write down each number n, we will update
4263.    the number of ones that have occurred and call this number f(n,1). The
4264.    first values for f(n,1), then, are as follows:
4265.
4266.                                    n  f(n,1)
4267.                                    0  0
4268.                                    1  1
4269.                                    2  1
4270.                                    3  1
4271.                                    4  1
4272.                                    5  1
4273.                                    6  1
4274.                                    7  1
4275.                                    8  1
4276.                                    9  1
4277.                                    10 2
4278.                                    11 4
4279.                                    12 5
4280.
4281.    Note that f(n,1) never equals 3.
4282.    So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The
4283.    next solution is n=199981.
4284.
4285.    In the same manner the function f(n,d) gives the total number of digits d
4286.    that have been written down after the number n has been written.
4287.    In fact, for every digit d ≠ 0, 0 is the first solution of the equation
4288.    f(n,d)=n.
4289.
4290.    Let s(d) be the sum of all the solutions for which f(n,d)=n.
4291.    You are given that s(1)=22786974071.
4292.
4293.    Find ∑ s(d) for 1 ≤ d ≤ 9.
4294.
4295.    Note: if, for some n, f(n,d)=n for more than one value of d this value of
4296.    n is counted again for every value of d for which f(n,d)=n.
4297.
4298.
4300.
4301.
4302. Problem 157
4303. ===========
4304.
4305.
4306.    Consider the diophantine equation ^1/[a]+^1/[b]= ^p/[10^n] with a, b, p, n
4307.    positive integers and a ≤ b.
4308.    For n=1 this equation has 20 solutions that are listed below:
4309.
4310. 1/1+1/1=20/10   1/1+1/2=15/10  1/1+1/5=12/10  1/1+1/10=11/10 1/2+1/2=10/10
4311. 1/2+1/5=7/10    1/2+1/10=6/10  1/3+1/6=5/10   1/3+1/15=4/10  1/4+1/4=5/10
4312. 1/4+1/20=3/10   1/5+1/5=4/10   1/5+1/10=3/10  1/6+1/30=2/10  1/10+1/10=2/10
4313. 1/11+1/110=1/10 1/12+1/60=1/10 1/14+1/35=1/10 1/15+1/30=1/10 1/20+1/20=1/10
4314.
4315.    How many solutions has this equation for 1 ≤ n ≤ 9?
4316.
4317.
4319.
4320.
4321. Problem 158
4322. ===========
4323.
4324.
4325.    Taking three different letters from the 26 letters of the alphabet,
4326.    character strings of length three can be formed.
4327.    Examples are 'abc', 'hat' and 'zyx'.
4328.    When we study these three examples we see that for 'abc' two characters
4329.    come lexicographically after its neighbour to the left.
4330.    For 'hat' there is exactly one character that comes lexicographically
4331.    after its neighbour to the left. For 'zyx' there are zero characters that
4332.    come lexicographically after its neighbour to the left.
4333.    In all there are 10400 strings of length 3 for which exactly one character
4334.    comes lexicographically after its neighbour to the left.
4335.
4336.    We now consider strings of n ≤ 26 different characters from the alphabet.
4337.    For every n, p(n) is the number of strings of length n for which exactly
4338.    one character comes lexicographically after its neighbour to the left.
4339.
4340.    What is the maximum value of p(n)?
4341.
4342.
4344.
4345.
4346. Problem 159
4347. ===========
4348.
4349.
4350.    A composite number can be factored many different ways. For instance, not
4351.    including multiplication by one, 24 can be factored in 7 distinct ways:
4352.
4353.    24 = 2x2x2x3
4354.    24 = 2x3x4
4355.    24 = 2x2x6
4356.    24 = 4x6
4357.    24 = 3x8
4358.    24 = 2x12
4359.    24 = 24
4360.
4361.    Recall that the digital root of a number, in base 10, is found by adding
4362.    together the digits of that number, and repeating that process until a
4363.    number is arrived at that is less than 10. Thus the digital root of 467 is
4364.    8.
4365.
4366.    We shall call a Digital Root Sum (DRS) the sum of the digital roots of the
4367.    individual factors of our number.
4368.    The chart below demonstrates all of the DRS values for 24.
4369.
4370.                         ┌─────────────┬────────────────┐
4371.                         │Factorisation│Digital Root Sum│
4372.                         ├─────────────┼────────────────┤
4373.                         │2x2x2x3      │       9        │
4374.                         ├─────────────┼────────────────┤
4375.                         │2x3x4        │       9        │
4376.                         ├─────────────┼────────────────┤
4377.                         │2x2x6        │       10       │
4378.                         ├─────────────┼────────────────┤
4379.                         │4x6          │       10       │
4380.                         ├─────────────┼────────────────┤
4381.                         │3x8          │       11       │
4382.                         ├─────────────┼────────────────┤
4383.                         │2x12         │       5        │
4384.                         ├─────────────┼────────────────┤
4385.                         │24           │       6        │
4386.                         └─────────────┴────────────────┘
4387.
4388.    The maximum Digital Root Sum of 24 is 11.
4389.    The function mdrs(n) gives the maximum Digital Root Sum of n. So
4390.    mdrs(24)=11.
4391.    Find ∑mdrs(n) for 1 < n < 1,000,000.
4392.
4393.
4395.
4396.
4397. Problem 160
4398. ===========
4399.
4400.
4401.    For any N, let f(N) be the last five digits before the trailing zeroes in
4402.    N!.
4403.    For example,
4404.
4405.    9! = 362880 so f(9)=36288
4406.    10! = 3628800 so f(10)=36288
4407.    20! = 2432902008176640000 so f(20)=17664
4408.
4409.    Find f(1,000,000,000,000)
4410.
4411.
4413.
4414.
4415. Problem 161
4416. ===========
4417.
4418.
4419.    A triomino is a shape consisting of three squares joined via the
4420.    edges.There are two basic forms:
4421.
4422.    If all possible orientations are taken into account there are six:
4423.
4424.    Any n by m grid for which nxm is divisible by 3 can be tiled with
4425.    triominoes.
4426.    If we consider tilings that can be obtained by reflection or rotation from
4427.    another tiling as different there are 41 ways a 2 by 9 grid can be tiled
4428.    with triominoes:
4429.
4430.    In how many ways can a 9 by 12 grid be tiled in this way by triominoes?
4431.
4432.
4433.    p_161_trio1.gif
4434.    p_161_trio3.gif
4435.    p_161_k9.gif
4437.
4438.
4439. Problem 162
4440. ===========
4441.
4442.
4443.    In the hexadecimal number system numbers are represented using 16
4444.    different digits:
4445.
4446.                         0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
4447.
4448.    The hexadecimal number AF when written in the decimal number system equals
4449.    10x16+15=175.
4450.
4451.    In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1
4452.    and A are all present.
4453.    Like numbers written in base ten we write hexadecimal numbers without
4455.
4457.    exist with all of the digits 0,1, and A present at least once?
4459.
4460.    (A,B,C,D,E and F in upper case, without any leading or trailing code that
4461.    marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and
4462.    not: 1a3f and not 0x1a3f and not 1A3F and not #1A3F and not 0000001A3F) 4463. 4464. 4465. Answer: 049419b9fdad9af74d5888626fff56a3 4466. 4467. 4468. Problem 163 4469. =========== 4470. 4471. 4472. Consider an equilateral triangle in which straight lines are drawn from 4473. each vertex to the middle of the opposite side, such as in the size 1 4474. triangle in the sketch below. 4475. 4476. Sixteen triangles of either different shape or size or orientation or 4477. location can now be observed in that triangle. Using size 1 triangles as 4478. building blocks, larger triangles can be formed, such as the size 2 4479. triangle in the above sketch. One-hundred and four triangles of either 4480. different shape or size or orientation or location can now be observed in 4481. that size 2 triangle. 4482. 4483. It can be observed that the size 2 triangle contains 4 size 1 triangle 4484. building blocks. A size 3 triangle would contain 9 size 1 triangle 4485. building blocks and a size n triangle would thus contain n^2 size 1 4486. triangle building blocks. 4487. 4488. If we denote T(n) as the number of triangles present in a triangle of size 4489. n, then 4490. 4491. T(1) = 16 4492. T(2) = 104 4493. 4494. Find T(36). 4495. 4496. 4497. p_163.gif 4498. Answer: 343047 4499. 4500. 4501. Problem 164 4502. =========== 4503. 4504. 4505. How many 20 digit numbers n (without any leading zero) exist such that no 4506. three consecutive digits of n have a sum greater than 9? 4507. 4508. 4509. Answer: 6e96debf3bfe7cc132401bafe5a5d6d6 4510. 4511. 4512. Problem 165 4513. =========== 4514. 4515. 4516. A segment is uniquely defined by its two endpoints. 4517. By considering two line segments in plane geometry there are three 4518. possibilities: 4519. the segments have zero points, one point, or infinitely many points in 4520. common. 4521. 4522. Moreover when two segments have exactly one point in common it might be 4523. the case that that common point is an endpoint of either one of the 4524. segments or of both. If a common point of two segments is not an endpoint 4525. of either of the segments it is an interior point of both segments. 4526. We will call a common point T of two segments L and L a true 4527. intersection point of L and L if T is the only common point of L 4528. and L and T is an interior point of both segments. 4529. 4530. Consider the three segments L, L, and L: 4531. 4532. L: (27, 44) to (12, 32) 4533. L: (46, 53) to (17, 62) 4534. L: (46, 70) to (22, 40) 4535. 4536. It can be verified that line segments L and L have a true 4537. intersection point. We note that as the one of the end points of L: 4538. (22,40) lies on L this is not considered to be a true point of 4539. intersection. L and L have no common point. So among the three line 4540. segments, we find one true intersection point. 4541. 4542. Now let us do the same for 5000 line segments. To this end, we generate 4543. 20000 numbers using the so-called "Blum Blum Shub" pseudo-random number 4544. generator. 4545. 4546. s = 290797 4547. 4548. s[n+1] = s[n]×s[n] (modulo 50515093) 4549. 4550. t[n] = s[n] (modulo 500) 4551. 4552. To create each line segment, we use four consecutive numbers t[n]. That 4553. is, the first line segment is given by: 4554. 4555. (t, t) to (t, t) 4556. 4557. The first four numbers computed according to the above generator should 4558. be: 27, 144, 12 and 232. The first segment would thus be (27,144) to 4559. (12,232). 4560. 4561. How many distinct true intersection points are found among the 5000 line 4562. segments? 4563. 4564. 4565. Answer: 2868868 4566. 4567. 4568. Problem 166 4569. =========== 4570. 4571. 4572. A 4x4 grid is filled with digits d, 0 ≤ d ≤ 9. 4573. 4574. It can be seen that in the grid 4575. 4576. 6 3 3 0 4577. 5 0 4 3 4578. 0 7 1 4 4579. 1 2 4 5 4580. 4581. the sum of each row and each column has the value 12. Moreover the sum of 4582. each diagonal is also 12. 4583. 4584. In how many ways can you fill a 4x4 grid with the digits d, 0 ≤ d ≤ 9 so 4585. that each row, each column, and both diagonals have the same sum? 4586. 4587. 4588. Answer: 7130034 4589. 4590. 4591. Problem 167 4592. =========== 4593. 4594. 4595. For two positive integers a and b, the Ulam sequence U(a,b) is defined by 4596. U(a,b) = a, U(a,b) = b and for k > 2,U(a,b)[k] is the smallest 4597. integer greater than U(a,b)[(k-1)] which can be written in exactly one way 4598. as the sum of two distinct previous members of U(a,b). 4599. 4600. For example, the sequence U(1,2) begins with 4601. 1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8; 4602. 5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations 4603. as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4. 4604. 4605. Find ∑U(2,2n+1)[k] for 2 ≤ n ≤10, where k = 10^11. 4606. 4607. 4608. Answer: aa5b61f6f4d96cbaeb5944b8fcdf64a3 4609. 4610. 4611. Problem 168 4612. =========== 4613. 4614. 4615. Consider the number 142857. We can right-rotate this number by moving the 4616. last digit (7) to the front of it, giving us 714285. 4617. It can be verified that 714285=5×142857. 4618. This demonstrates an unusual property of 142857: it is a divisor of its 4619. right-rotation. 4620. 4621. Find the last 5 digits of the sum of all integers n, 10 < n < 10^100, that 4622. have this property. 4623. 4624. 4625. Answer: 59206 4626. 4627. 4628. Problem 169 4629. =========== 4630. 4631. 4632. Define f(0)=1 and f(n) to be the number of different ways n can be 4633. expressed as a sum of integer powers of 2 using each power no more than 4634. twice. 4635. 4636. For example, f(10)=5 since there are five different ways to express 10: 4637. 4638. 1 + 1 + 8 4639. 1 + 1 + 4 + 4 4640. 1 + 1 + 2 + 2 + 4 4641. 2 + 4 + 4 4642. 2 + 8 4643. 4644. What is f(10^25)? 4645. 4646. 4647. Answer: d149d4836703a8908becea56ddd3ed42 4648. 4649. 4650. Problem 170 4651. =========== 4652. 4653. 4654. Take the number 6 and multiply it by each of 1273 and 9854: 4655. 4656. 6 × 1273 = 7638 4657. 6 × 9854 = 59124 4658. 4659. By concatenating these products we get the 1 to 9 pandigital 763859124. We 4660. will call 763859124 the "concatenated product of 6 and (1273,9854)". 4661. Notice too, that the concatenation of the input numbers, 612739854, is 4662. also 1 to 9 pandigital. 4663. 4664. The same can be done for 0 to 9 pandigital numbers. 4665. 4666. What is the largest 0 to 9 pandigital 10-digit concatenated product of an 4667. integer with two or more other integers, such that the concatenation of 4668. the input numbers is also a 0 to 9 pandigital 10-digit number? 4669. 4670. 4671. Answer: 6ffe65352f717c1731666a107ace96c1 4672. 4673. 4674. Problem 171 4675. =========== 4676. 4677. 4678. For a positive integer n, let f(n) be the sum of the squares of the digits 4679. (in base 10) of n, e.g. 4680. 4681. f(3) = 3^2 = 9, 4682. f(25) = 2^2 + 5^2 = 4 + 25 = 29, 4683. f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 4684. 4685. Find the last nine digits of the sum of all n, 0 < n < 10^20, such that 4686. f(n) is a perfect square. 4687. 4688. 4689. Answer: 142989277 4690. 4691. 4692. Problem 172 4693. =========== 4694. 4695. 4696. How many 18-digit numbers n (without leading zeros) are there such that no 4697. digit occurs more than three times in n? 4698. 4699. 4700. Answer: f5f260ee21ead7478403c2ccd18a1829 4701. 4702. 4703. Problem 173 4704. =========== 4705. 4706. 4707. We shall define a square lamina to be a square outline with a square 4708. "hole" so that the shape possesses vertical and horizontal symmetry. For 4709. example, using exactly thirty-two square tiles we can form two different 4710. square laminae: 4711. 4712. With one-hundred tiles, and not necessarily using all of the tiles at one 4713. time, it is possible to form forty-one different square laminae. 4714. 4715. Using up to one million tiles how many different square laminae can be 4716. formed? 4717. 4718. 4719. p_173_square_laminas.gif 4720. Answer: 1572729 4721. 4722. 4723. Problem 174 4724. =========== 4725. 4726. 4727. We shall define a square lamina to be a square outline with a square 4728. "hole" so that the shape possesses vertical and horizontal symmetry. 4729. 4730. Given eight tiles it is possible to form a lamina in only one way: 3x3 4731. square with a 1x1 hole in the middle. However, using thirty-two tiles it 4732. is possible to form two distinct laminae. 4733. 4734. If t represents the number of tiles used, we shall say that t = 8 is type 4735. L(1) and t = 32 is type L(2). 4736. 4737. Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for 4738. example, N(15) = 832. 4739. 4740. What is ∑ N(n) for 1 ≤ n ≤ 10? 4741. 4742. 4743. p_173_square_laminas.gif 4744. Answer: 209566 4745. 4746. 4747. Problem 175 4748. =========== 4749. 4750. Define f(0)=1 and f(n) to be the number of ways to write n as a sum of 4751. powers of 2 where no power occurs more than twice. 4752. 4753. For example, f(10)=5 since there are five different ways to express 10: 4754. 10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1 4755. 4756. It can be shown that for every fraction p/q (p>0, q>0) there exists at 4757. least one integer n such that 4758. f(n)/f(n-1)=p/q. 4759. 4760. For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. 4761. The binary expansion of 241 is 11110001. 4762. Reading this binary number from the most significant bit to the least 4763. significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the 4764. string 4,3,1 the Shortened Binary Expansion of 241. 4765. 4766. Find the Shortened Binary Expansion of the smallest n for which 4767. f(n)/f(n-1)=123456789/987654321. 4768. 4769. Give your answer as comma separated integers, without any whitespaces. 4770. 4771. Answer: 796dddd004c3465229058072f5b4583e 4772. 4773. 4774. Problem 176 4775. =========== 4776. 4777. 4778. The four right-angled triangles with sides (9,12,15), (12,16,20), 4779. (5,12,13) and (12,35,37) all have one of the shorter sides (catheti) equal 4780. to 12. It can be shown that no other integer sided right-angled triangle 4781. exists with one of the catheti equal to 12. 4782. 4783. Find the smallest integer that can be the length of a cathetus of exactly 4784. 47547 different integer sided right-angled triangles. 4785. 4786. 4787. Answer: c47c782ebaf8cdbb60eebfa86cd0003c 4788. 4789. 4790. Problem 177 4791. =========== 4792. 4793. 4794. Let ABCD be a convex quadrilateral, with diagonals AC and BD. At each 4795. vertex the diagonal makes an angle with each of the two sides, creating 4796. eight corner angles. 4797. 4798. For example, at vertex A, the two angles are CAD, CAB. 4799. 4800. We call such a quadrilateral for which all eight corner angles have 4801. integer values when measured in degrees an "integer angled quadrilateral". 4802. An example of an integer angled quadrilateral is a square, where all eight 4803. corner angles are 45°. Another example is given by DAC = 20°, BAC = 60°, 4804. ABD = 50°, CBD = 30°, BCA = 40°, DCA = 30°, CDB = 80°, ADB = 50°. 4805. 4806. What is the total number of non-similar integer angled quadrilaterals? 4807. 4808. Note: In your calculations you may assume that a calculated angle is 4809. integral if it is within a tolerance of 10^-9 of an integer value. 4810. 4811. 4812. p_177_quad.gif 4813. Answer: 129325 4814. 4815. 4816. Problem 178 4817. =========== 4818. 4819. Consider the number 45656. 4820. It can be seen that each pair of consecutive digits of 45656 has a 4821. difference of one. 4822. A number for which every pair of consecutive digits has a difference of 4823. one is called a step number. 4824. A pandigital number contains every decimal digit from 0 to 9 at least 4825. once. 4826. How many pandigital step numbers less than 10^40 are there? 4827. 4828. Answer: 2ffddfa898fa5df6321aebea84d4f33f 4829. 4830. 4831. Problem 179 4832. =========== 4833. 4834. 4835. Find the number of integers 1 < n < 10^7, for which n and n + 1 have the 4836. same number of positive divisors. For example, 14 has the positive 4837. divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15. 4838. 4839. 4840. Answer: 986262 4841. 4842. 4843. Problem 180 4844. =========== 4845. 4846. 4847. For any integer n, consider the three functions 4848. 4849. f[1,n](x,y,z) = x^n+1 + y^n+1 − z^n+1 4850. f[2,n](x,y,z) = (xy + yz + zx)*(x^n-1 + y^n-1 − z^n-1) 4851. f[3,n](x,y,z) = xyz*(x^n-2 + y^n-2 − z^n-2) 4852. 4853. and their combination 4854. 4855. f[n](x,y,z) = f[1,n](x,y,z) + f[2,n](x,y,z) − f[3,n](x,y,z) 4856. 4857. We call (x,y,z) a golden triple of order k if x, y, and z are all rational 4858. numbers of the form a / b with 4859. 0 < a < b ≤ k and there is (at least) one integer n, so that f[n](x,y,z) = 4860. 0. 4861. 4862. Let s(x,y,z) = x + y + z. 4863. Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples 4864. (x,y,z) of order 35. 4865. All the s(x,y,z) and t must be in reduced form. 4866. 4867. Find u + v. 4868. 4869. 4870. Answer: 6459f69d151314c59df404868f45fa96 4871. 4872. 4873. Problem 181 4874. =========== 4875. 4876. 4877. Having three black objects B and one white object W they can be grouped in 4878. 7 ways like this: 4879. 4880. (BBBW) (B,BBW) (B,B,BW) (B,B,B,W) (B,BB,W) (BBB,W) (BB,BW) 4881. 4882. In how many ways can sixty black objects B and forty white objects W be 4883. thus grouped? 4884. 4885. 4886. Answer: 0e1233ecbc058dabf54a8602eac55d95 4887. 4888. 4889. Problem 182 4890. =========== 4891. 4892. 4893. The RSA encryption is based on the following procedure: 4894. 4895. Generate two distinct primes p and q. 4896. Compute n=pq and φ=(p-1)(q-1). 4897. Find an integer e, 1<e<φ, such that gcd(e,φ)=1. 4898. 4899. A message in this system is a number in the interval [0,n-1]. 4900. A text to be encrypted is then somehow converted to messages (numbers in 4901. the interval [0,n-1]). 4902. To encrypt the text, for each message, m, c=m^e mod n is calculated. 4903. 4904. To decrypt the text, the following procedure is needed: calculate d such 4905. that ed=1 mod φ, then for each encrypted message, c, calculate m=c^d mod 4906. n. 4907. 4908. There exist values of e and m such that m^e mod n=m. 4909. We call messages m for which m^e mod n=m unconcealed messages. 4910. 4911. An issue when choosing e is that there should not be too many unconcealed 4912. messages. 4913. For instance, let p=19 and q=37. 4914. Then n=19*37=703 and φ=18*36=648. 4915. If we choose e=181, then, although gcd(181,648)=1 it turns out that all 4916. possible messages 4917. m (0≤m≤n-1) are unconcealed when calculating m^e mod n. 4918. For any valid choice of e there exist some unconcealed messages. 4919. It's important that the number of unconcealed messages is at a minimum. 4920. 4921. Choose p=1009 and q=3643. 4922. Find the sum of all values of e, 1<e<φ(1009,3643) and gcd(e,φ)=1, so that 4923. the number of unconcealed messages for this value of e is at a minimum. 4924. 4925. 4926. Answer: 088ad9a61e60b9309e91cfc3ed27d729 4927. 4928. 4929. Problem 183 4930. =========== 4931. 4932. 4933. Let N be a positive integer and let N be split into k equal parts, r = 4934. N/k, so that N = r + r + ... + r. 4935. Let P be the product of these parts, P = r × r × ... × r = r^k. 4936. 4937. For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 4938. 2.2 + 2.2, then P = 2.2^5 = 51.53632. 4939. 4940. Let M(N) = P[max] for a given value of N. 4941. 4942. It turns out that the maximum for N = 11 is found by splitting eleven into 4943. four equal parts which leads to P[max] = (11/4)^4; that is, M(11) = 4944. 14641/256 = 57.19140625, which is a terminating decimal. 4945. 4946. However, for N = 8 the maximum is achieved by splitting it into three 4947. equal parts, so M(8) = 512/27, which is a non-terminating decimal. 4948. 4949. Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is 4950. a terminating decimal. 4951. 4952. For example, ΣD(N) for 5 ≤ N ≤ 100 is 2438. 4953. 4954. Find ΣD(N) for 5 ≤ N ≤ 10000. 4955. 4956. 4957. Answer: 48861552 4958. 4959. 4960. Problem 184 4961. =========== 4962. 4963. 4964. Consider the set I[r] of points (x,y) with integer co-ordinates in the 4965. interior of the circle with radius r, centered at the origin, i.e. x^2 + 4966. y^2 < r^2. 4967. 4968. For a radius of 2, I contains the nine points (0,0), (1,0), (1,1), 4969. (0,1), (-1,1), (-1,0), (-1,-1), (0,-1) and (1,-1). There are eight 4970. triangles having all three vertices in I which contain the origin in 4971. the interior. Two of them are shown below, the others are obtained from 4972. these by rotation. 4973. 4974. For a radius of 3, there are 360 triangles containing the origin in the 4975. interior and having all vertices in I and for I the number is 10600. 4976. 4977. How many triangles are there containing the origin in the interior and 4978. having all three vertices in I? 4979. 4980. 4981. p_184.gif 4982. Answer: aa80f8619ed594e5d7852564457dbca6 4983. 4984. 4985. Problem 185 4986. =========== 4987. 4988. 4989. The game Number Mind is a variant of the well known game Master Mind. 4990. 4991. Instead of coloured pegs, you have to guess a secret sequence of digits. 4992. After each guess you're only told in how many places you've guessed the 4993. correct digit. So, if the sequence was 1234 and you guessed 2036, you'd be 4994. told that you have one correct digit; however, you would NOT be told that 4995. you also have another digit in the wrong place. 4996. 4997. For instance, given the following guesses for a 5-digit secret sequence, 4998. 4999. 90342 ;2 correct 5000. 70794 ;0 correct 5001. 39458 ;2 correct 5002. 34109 ;1 correct 5003. 51545 ;2 correct 5004. 12531 ;1 correct 5005. 5006. The correct sequence 39542 is unique. 5007. 5008. Based on the following guesses, 5009. 5010. 5616185650518293 ;2 correct 5011. 3847439647293047 ;1 correct 5012. 5855462940810587 ;3 correct 5013. 9742855507068353 ;3 correct 5014. 4296849643607543 ;3 correct 5015. 3174248439465858 ;1 correct 5016. 4513559094146117 ;2 correct 5017. 7890971548908067 ;3 correct 5018. 8157356344118483 ;1 correct 5019. 2615250744386899 ;2 correct 5020. 8690095851526254 ;3 correct 5021. 6375711915077050 ;1 correct 5022. 6913859173121360 ;1 correct 5023. 6442889055042768 ;2 correct 5024. 2321386104303845 ;0 correct 5025. 2326509471271448 ;2 correct 5026. 5251583379644322 ;2 correct 5027. 1748270476758276 ;3 correct 5028. 4895722652190306 ;1 correct 5029. 3041631117224635 ;3 correct 5030. 1841236454324589 ;3 correct 5031. 2659862637316867 ;2 correct 5032. 5033. Find the unique 16-digit secret sequence. 5034. 5035. 5036. Answer: 70f84864f21c4bf07ee53436580cd4bb 5037. 5038. 5039. Problem 186 5040. =========== 5041. 5042. 5043. Here are the records from a busy telephone system with one million users: 5044. 5045. ┌─────────────────┬─────────┬─────────┐ 5046. │RecNr │ Caller │ Called │ 5047. ├─────────────────┼─────────┼─────────┤ 5048. │ 1 │ 200007 │ 100053 │ 5049. ├─────────────────┼─────────┼─────────┤ 5050. │ 2 │ 600183 │ 500439 │ 5051. ├─────────────────┼─────────┼─────────┤ 5052. │ 3 │ 600863 │ 701497 │ 5053. ├─────────────────┼─────────┼─────────┤ 5054. │ ... │ ... │ ... │ 5055. └─────────────────┴─────────┴─────────┘ 5056. 5057. The telephone number of the caller and the called number in record n are 5058. Caller(n) = S[2n-1] and Called(n) = S[2n] where S[1,2,3,...] come from the 5059. "Lagged Fibonacci Generator": 5060. 5061. For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3] (modulo 1000000) 5062. For 56 ≤ k, S[k] = [S[k-24] + S[k-55]] (modulo 1000000) 5063. 5064. If Caller(n) = Called(n) then the user is assumed to have misdialled and 5065. the call fails; otherwise the call is successful. 5066. 5067. From the start of the records, we say that any pair of users X and Y are 5068. friends if X calls Y or vice-versa. Similarly, X is a friend of a friend 5069. of Z if X is a friend of Y and Y is a friend of Z; and so on for longer 5070. chains. 5071. 5072. The Prime Minister's phone number is 524287. After how many successful 5073. calls, not counting misdials, will 99% of the users (including the PM) be 5074. a friend, or a friend of a friend etc., of the Prime Minister? 5075. 5076. 5077. Answer: 2325629 5078. 5079. 5080. Problem 187 5081. =========== 5082. 5083. 5084. A composite is a number containing at least two prime factors. For 5085. example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3. 5086. 5087. There are ten composites below thirty containing precisely two, not 5088. necessarily distinct, prime factors:4, 6, 9, 10, 14, 15, 21, 22, 25, 26. 5089. 5090. How many composite integers, n < 10^8, have precisely two, not necessarily 5091. distinct, prime factors? 5092. 5093. 5094. Answer: 17427258 5095. 5096. 5097. Problem 188 5098. =========== 5099. 5100. 5101. The hyperexponentiation or tetration of a number a by a positive integer 5102. b, denoted by a↑↑b or ^ba, is recursively defined by: 5103. 5104. a↑↑1 = a, 5105. a↑↑(k+1) = a^(a↑↑k). 5106. 5107. Thus we have e.g. 3↑↑2 = 3^3 = 27, hence 3↑↑3 = 3^27 = 7625597484987 and 5108. 3↑↑4 is roughly 10^3.6383346400240996*10^12. 5109. 5110. Find the last 8 digits of 1777↑↑1855. 5111. 5112. 5113. Answer: 95962097 5114. 5115. 5116. Problem 189 5117. =========== 5118. 5119. 5120. Consider the following configuration of 64 triangles: 5121. 5122. We wish to colour the interior of each triangle with one of three colours: 5123. red, green or blue, so that no two neighbouring triangles have the same 5124. colour. Such a colouring shall be called valid. Here, two triangles are 5125. said to be neighbouring if they share an edge. 5126. Note: if they only share a vertex, then they are not neighbours. 5127. 5128. For example, here is a valid colouring of the above grid: 5129. 5130. A colouring C' which is obtained from a colouring C by rotation or 5131. reflection is considered distinct from C unless the two are identical. 5132. 5133. How many distinct valid colourings are there for the above configuration? 5134. 5135. 5136. p_189_grid.gif 5137. p_189_colours.gif 5138. Answer: d3dfdd37601678212b746c34699f1484 5139. 5140. 5141. Problem 190 5142. =========== 5143. 5144. 5145. Let S[m] = (x, x, ... , x[m]) be the m-tuple of positive real 5146. numbers with x + x + ... + x[m] = m for which P[m] = x * x^2 * 5147. ... * x[m]^m is maximised. 5148. 5149. For example, it can be verified that [P] = 4112 ([ ] is the integer 5150. part function). 5151. 5152. Find Σ[P[m]] for 2 ≤ m ≤ 15. 5153. 5154. 5155. Answer: 40cfcabd9b30d79ec81151fc756e9946 5156. 5157. 5158. Problem 191 5159. =========== 5160. 5161. 5162. A particular school offers cash rewards to children with good attendance 5163. and punctuality. If they are absent for three consecutive days or late on 5164. more than one occasion then they forfeit their prize. 5165. 5166. During an n-day period a trinary string is formed for each child 5167. consisting of L's (late), O's (on time), and A's (absent). 5168. 5169. Although there are eighty-one trinary strings for a 4-day period that can 5170. be formed, exactly forty-three strings would lead to a prize: 5171. 5172. OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA 5173. OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO 5174. AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL 5175. AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA 5176. LAOO LAOA LAAO 5177. 5178. How many "prize" strings exist over a 30-day period? 5179. 5180. 5181. Answer: e04dfa598b22a87570f63063f3ff595d 5182. 5183. 5184. Problem 192 5185. =========== 5186. 5187. 5188. Let x be a real number. 5189. A best approximation to x for the denominator bound d is a rational number 5190. r/s in reduced form, with s ≤ d, such that any rational number which is 5191. closer to x than r/s has a denominator larger than d: 5192. 5193. |p/q-x| < |r/s-x| ⇒ q > d 5194. 5195. For example, the best approximation to √13 for the denominator bound 20 is 5196. 18/5 and the best approximation to √13 for the denominator bound 30 is 5197. 101/28. 5198. 5199. Find the sum of all denominators of the best approximations to √n for the 5200. denominator bound 10^12, where n is not a perfect square and 1 < n ≤ 5201. 100000. 5202. 5203. 5204. Answer: e5ec7d4b094709b1fcebbd73b10e6264 5205. 5206. 5207. Problem 193 5208. =========== 5209. 5210. 5211. A positive integer n is called squarefree, if no square of a prime divides 5212. n, thus 1, 2, 3, 5, 6, 7, 10, 11 are squarefree, but not 4, 8, 9, 12. 5213. 5214. How many squarefree numbers are there below 2^50? 5215. 5216. 5217. Answer: ea29fcf755b560777b0b6d8714234d18 5218. 5219. 5220. Problem 194 5221. =========== 5222. 5223. 5224. Consider graphs built with the units A: and B: , where the units are glued 5225. alongthe vertical edges as in the graph . 5226. 5227. A configuration of type (a,b,c) is a graph thus built of a units A and b 5228. units B, where the graph's vertices are coloured using up to c colours, so 5229. that no two adjacent vertices have the same colour. 5230. The compound graph above is an example of a configuration of type (2,2,6), 5231. in fact of type (2,2,c) for all c ≥ 4. 5232. 5233. Let N(a,b,c) be the number of configurations of type (a,b,c). 5234. For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736. 5235. 5236. Find the last 8 digits of N(25,75,1984). 5237. 5238. 5239. p_194_GraphA.png 5240. p_194_GraphB.png 5241. p_194_Fig.png 5242. Answer: 61190912 5243. 5244. 5245. Problem 195 5246. =========== 5247. 5248. 5249. Let's call an integer sided triangle with exactly one angle of 60 degrees 5250. a 60-degree triangle. 5251. Let r be the radius of the inscribed circle of such a 60-degree triangle. 5252. 5253. There are 1234 60-degree triangles for which r ≤ 100. 5254. Let T(n) be the number of 60-degree triangles for which r ≤ n, so 5255. T(100) = 1234, T(1000) = 22767, and T(10000) = 359912. 5256. 5257. Find T(1053779). 5258. 5259. 5260. Answer: 75085391 5261. 5262. 5263. Problem 196 5264. =========== 5265. 5266. 5267. Build a triangle from all positive integers in the following way: 5268. 5269. 1 5270. 2 3 5271. 4 5 6 5272. 7 8 9 10 5273. 11 12 13 14 15 5274. 16 17 18 19 20 21 5275. 22 23 24 25 26 27 28 5276. 29 30 31 32 33 34 35 36 5277. 37 38 39 40 41 42 43 44 45 5278. 46 47 48 49 50 51 52 53 54 55 5279. 56 57 58 59 60 61 62 63 64 65 66 5280. . . . 5281. 5282. Each positive integer has up to eight neighbours in the triangle. 5283. 5284. A set of three primes is called a prime triplet if one of the three primes 5285. has the other two as neighbours in the triangle. 5286. 5287. For example, in the second row, the prime numbers 2 and 3 are elements of 5288. some prime triplet. 5289. 5290. If row 8 is considered, it contains two primes which are elements of some 5291. prime triplet, i.e. 29 and 31. 5292. If row 9 is considered, it contains only one prime which is an element of 5293. some prime triplet: 37. 5294. 5295. Define S(n) as the sum of the primes in row n which are elements of any 5296. prime triplet. 5297. Then S(8)=60 and S(9)=37. 5298. 5299. You are given that S(10000)=950007619. 5300. 5301. Find S(5678027) + S(7208785). 5302. 5303. 5304. Answer: fb6b6b0a4b7b31ba429152bc0b6bd037 5305. 5306. 5307. Problem 197 5308. =========== 5309. 5310. 5311. Given is the function f(x) = ⌊2^30.403243784-x^2⌋ × 10^-9 ( ⌊ ⌋ is the 5312. floor-function), 5313. the sequence u[n] is defined by u = -1 and u[n+1] = f(u[n]). 5314. 5315. Find u[n] + u[n+1] for n = 10^12. 5316. Give your answer with 9 digits after the decimal point. 5317. 5318. 5319. Answer: c98cbf87636906f2465d481be815e454 5320. 5321. 5322. Problem 198 5323. =========== 5324. 5325. 5326. A best approximation to a real number x for the denominator bound d is a 5327. rational number r/s (in reduced form) with s ≤ d, so that any rational 5328. number p/q which is closer to x than r/s has q > d. 5329. 5330. Usually the best approximation to a real number is uniquely determined for 5331. all denominator bounds. However, there are some exceptions, e.g. 9/40 has 5332. the two best approximations 1/4 and 1/5 for the denominator bound 6.We 5333. shall call a real number x ambiguous, if there is at least one denominator 5334. bound for which x possesses two best approximations. Clearly, an ambiguous 5335. number is necessarily rational. 5336. 5337. How many ambiguous numbers x = p/q,0 < x < 1/100, are there whose 5338. denominator q does not exceed 10^8? 5339. 5340. 5341. Answer: 52374425 5342. 5343. 5344. Problem 199 5345. =========== 5346. 5347. 5348. Three circles of equal radius are placed inside a larger circle such that 5349. each pair of circles is tangent to one another and the inner circles do 5350. not overlap. There are four uncovered "gaps" which are to be filled 5351. iteratively with more tangent circles. 5352. 5353. At each iteration, a maximally sized circle is placed in each gap, which 5354. creates more gaps for the next iteration. After 3 iterations (pictured), 5355. there are 108 gaps and the fraction of the area which is not covered by 5356. circles is 0.06790342, rounded to eight decimal places. 5357. 5358. What fraction of the area is not covered by circles after 10 iterations? 5359. Give your answer rounded to eight decimal places using the format 5360. x.xxxxxxxx . 5361. 5362. 5363. p_199_circles_in_circles.gif 5364. Answer: 0f8fd87159c28ae5fea6ac91a95d48dd 5365. 5366. 5367. Problem 200 5368. =========== 5369. 5370. 5371. We shall define a sqube to be a number of the form, p^2q^3, where p and q 5372. are distinct primes. 5373. For example, 200 = 5^22^3 or 120072949 = 23^261^3. 5374. 5375. The first five squbes are 72, 108, 200, 392, and 500. 5376. 5377. Interestingly, 200 is also the first number for which you cannot change 5378. any single digit to make a prime; we shall call such numbers, prime-proof. 5379. The next prime-proof sqube which contains the contiguous sub-string "200" 5380. is 1992008. 5381. 5382. Find the 200th prime-proof sqube containing the contiguous sub-string 5383. "200". 5384. 5385. 5386. Answer: c911c8e346aa813da5f5ed4f8e9128d8 5387. 5388. 5389. Problem 201 5390. =========== 5391. 5392. 5393. For any set A of numbers, let sum(A) be the sum of the elements of A. 5394. Consider the set B = {1,3,6,8,10,11}. 5395. There are 20 subsets of B containing three elements, and their sums are: 5396. 5397. sum({1,3,6}) = 10, 5398. sum({1,3,8}) = 12, 5399. sum({1,3,10}) = 14, 5400. sum({1,3,11}) = 15, 5401. sum({1,6,8}) = 15, 5402. sum({1,6,10}) = 17, 5403. sum({1,6,11}) = 18, 5404. sum({1,8,10}) = 19, 5405. sum({1,8,11}) = 20, 5406. sum({1,10,11}) = 22, 5407. sum({3,6,8}) = 17, 5408. sum({3,6,10}) = 19, 5409. sum({3,6,11}) = 20, 5410. sum({3,8,10}) = 21, 5411. sum({3,8,11}) = 22, 5412. sum({3,10,11}) = 24, 5413. sum({6,8,10}) = 24, 5414. sum({6,8,11}) = 25, 5415. sum({6,10,11}) = 27, 5416. sum({8,10,11}) = 29. 5417. 5418. Some of these sums occur more than once, others are unique. 5419. For a set A, let U(A,k) be the set of unique sums of k-element subsets of 5420. A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and 5421. sum(U(B,3)) = 156. 5422. 5423. Now consider the 100-element set S = {1^2, 2^2, ... , 100^2}. 5424. S has 100891344545564193334812497256 50-element subsets. 5425. 5426. Determine the sum of all integers which are the sum of exactly one of the 5427. 50-element subsets of S, i.e. find sum(U(S,50)). 5428. 5429. 5430. Answer: 115039000 5431. 5432. 5433. Problem 202 5434. =========== 5435. 5436. 5437. Three mirrors are arranged in the shape of an equilateral triangle, with 5438. their reflective surfaces pointing inwards. There is an infinitesimal gap 5439. at each vertex of the triangle through which a laser beam may pass. 5440. 5441. Label the vertices A, B and C. There are 2 ways in which a laser beam may 5442. enter vertex C, bounce off 11 surfaces, then exit through the same vertex: 5443. one way is shown below; the other is the reverse of that. 5444. 5445. There are 80840 ways in which a laser beam may enter vertex C, bounce off 5446. 1000001 surfaces, then exit through the same vertex. 5447. 5448. In how many ways can a laser beam enter at vertex C, bounce off 5449. 12017639147 surfaces, then exit through the same vertex? 5450. 5451. 5452. p_201_laserbeam.gif 5453. Answer: e9774949b5efad0d40d60ede379c5321 5454. 5455. 5456. Problem 203 5457. =========== 5458. 5459. 5460. The binomial coefficients ^nC[k] can be arranged in triangular form, 5461. Pascal's triangle, like this: 5462. 5463. 1 5464. 1 1 5465. 1 2 1 5466. 1 3 3 1 5467. 1 4 6 4 1 5468. 1 5 10 10 5 1 5469. 1 6 15 20 15 6 1 5470. 1 7 21 35 35 21 7 1 5471. 5472. ......... 5473. 5474. It can be seen that the first eight rows of Pascal's triangle contain 5475. twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35. 5476. 5477. A positive integer n is called squarefree if no square of a prime divides 5478. n.Of the twelve distinct numbers in the first eight rows of Pascal's 5479. triangle, all except 4 and 20 are squarefree.The sum of the distinct 5480. squarefree numbers in the first eight rows is 105. 5481. 5482. Find the sum of the distinct squarefree numbers in the first 51 rows of 5483. Pascal's triangle. 5484. 5485. 5486. Answer: d7ec16d216c923d3c927f46cfc914e92 5487. 5488. 5489. Problem 204 5490. =========== 5491. 5492. 5493. A Hamming number is a positive number which has no prime factor larger 5494. than 5. 5495. So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15. 5496. There are 1105 Hamming numbers not exceeding 10^8. 5497. 5498. We will call a positive number a generalised Hamming number of type n, if 5499. it has no prime factor larger than n. 5500. Hence the Hamming numbers are the generalised Hamming numbers of type 5. 5501. 5502. How many generalised Hamming numbers of type 100 are there which don't 5503. exceed 10^9? 5504. 5505. 5506. Answer: 2944730 5507. 5508. 5509. Problem 205 5510. =========== 5511. 5512. 5513. Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 5514. 3, 4. 5515. Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5516. 5, 6. 5517. 5518. Peter and Colin roll their dice and compare totals: the highest total 5519. wins. The result is a draw if the totals are equal. 5520. 5521. What is the probability that Pyramidal Pete beats Cubic Colin? Give your 5522. answer rounded to seven decimal places in the form 0.abcdefg 5523. 5524. 5525. Answer: ba6c6c3888227a0799eca38191b587be 5526. 5527. 5528. Problem 206 5529. =========== 5530. 5531. 5532. Find the unique positive integer whose square has the form 5533. 1_2_3_4_5_6_7_8_9_0, 5534. where each “_” is a single digit. 5535. 5536. 5537. Answer: 09f9d87cb4b1ebb34e1f607e55a351d8 5538. 5539. 5540. Problem 207 5541. =========== 5542. 5543. 5544. For some positive integers k, there exists an integer partition of the 5545. form 4^t = 2^t + k, 5546. where 4^t, 2^t, and k are all positive integers and t is a real number. 5547. 5548. The first two such partitions are 4^1 = 2^1 + 2 and 4^1.5849625... = 5549. 2^1.5849625... + 6. 5550. 5551. Partitions where t is also an integer are called perfect. 5552. For any m ≥ 1 let P(m) be the proportion of such partitions that are 5553. perfect with k ≤ m. 5554. Thus P(6) = 1/2. 5555. 5556. In the following table are listed some values of P(m) 5557. 5558. P(5) = 1/1 5559. P(10) = 1/2 5560. P(15) = 2/3 5561. P(20) = 1/2 5562. P(25) = 1/2 5563. P(30) = 2/5 5564. ... 5565. P(180) = 1/4 5566. P(185) = 3/13 5567. 5568. Find the smallest m for which P(m) < 1/12345 5569. 5570. 5571. Answer: 3f17b264ed1717fe5fbde1e399bd501f 5572. 5573. 5574. Problem 208 5575. =========== 5576. 5577. 5578. A robot moves in a series of one-fifth circular arcs (72°), with a free 5579. choice of a clockwise or an anticlockwise arc for each step, but no 5580. turning on the spot. 5581. 5582. One of 70932 possible closed paths of 25 arcs starting northward is 5583. 5584. Given that the robot starts facing North, how many journeys of 70 arcs in 5585. length can it take that return it, after the final arc, to its starting 5586. position? 5587. (Any arc may be traversed multiple times.) 5588. 5589. 5590. p_208_robotwalk.gif 5591. Answer: 3010e33173f30e0aac79e84835b48823 5592. 5593. 5594. Problem 209 5595. =========== 5596. 5597. 5598. A k-input binary truth table is a map from k input bits(binary digits, 0 5599. [false] or 1 [true]) to 1 output bit. For example, the 2-input binary 5600. truth tables for the logical AND and XOR functions are: 5601. 5602. ┌────┬────┬─────────┐ 5603. │ x │ y │x AND y │ 5604. ├────┼────┼─────────┤ 5605. │ 0 │ 0 │ 0 │ 5606. ├────┼────┼─────────┤ 5607. │ 0 │ 1 │ 0 │ 5608. ├────┼────┼─────────┤ 5609. │ 1 │ 0 │ 0 │ 5610. ├────┼────┼─────────┤ 5611. │ 1 │ 1 │ 1 │ 5612. └────┴────┴─────────┘ 5613. 5614. ┌────┬────┬─────────┐ 5615. │ x │ y │x XOR y │ 5616. ├────┼────┼─────────┤ 5617. │ 0 │ 0 │ 0 │ 5618. ├────┼────┼─────────┤ 5619. │ 0 │ 1 │ 1 │ 5620. ├────┼────┼─────────┤ 5621. │ 1 │ 0 │ 1 │ 5622. ├────┼────┼─────────┤ 5623. │ 1 │ 1 │ 0 │ 5624. └────┴────┴─────────┘ 5625. 5626. How many 6-input binary truth tables, τ, satisfy the formula 5627. 5628. τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0 5629. 5630. Answer: 954157aa4762df2ee29580ee5a351b13 5631. 5632. 5633. Problem 210 5634. =========== 5635. 5636. Consider the set S(r) of points (x,y) with integer coordinates satisfying 5637. |x| + |y| ≤ r. 5638. Let O be the point (0,0) and C the point (r/4,r/4). 5639. Let N(r) be the number of points B in S(r), so that the triangle OBC has 5640. an obtuse angle, i.e. the largest angle α satisfies 90°<α<180°. 5641. So, for example, N(4)=24 and N(8)=100. 5642. 5643. What is N(1,000,000,000)? 5644. 5645. 5646. Answer: 0c808b02789c4db462322ab2ac070bbb 5647. 5648. 5649. Problem 211 5650. =========== 5651. 5652. 5653. For a positive integer n, let σ(n) be the sum of the squares of its 5654. divisors. For example, 5655. 5656. σ(10) = 1 + 4 + 25 + 100 = 130. 5657. 5658. Find the sum of all n, 0 < n < 64,000,000 such that σ(n) is a perfect 5659. square. 5660. 5661. 5662. Answer: 5fe0ed146690e7bca448687a94353a73 5663. 5664. 5665. Problem 212 5666. =========== 5667. 5668. 5669. An axis-aligned cuboid, specified by parameters { (x,y,z), 5670. (dx,dy,dz) }, consists of all points (X,Y,Z) such that x ≤ X ≤ x+dx, 5671. y ≤ Y ≤ y+dy and z ≤ Z ≤ z+dz. The volume of the cuboid is the 5672. product, dx × dy × dz. The combined volume of a collection of cuboids is 5673. the volume of their union and will be less than the sum of the individual 5674. volumes if any cuboids overlap. 5675. 5676. Let C,...,C be a collection of 50000 axis-aligned cuboids such 5677. that C[n] has parameters 5678. 5679. x = S[6n-5] modulo 10000 5680. y = S[6n-4] modulo 10000 5681. z = S[6n-3] modulo 10000 5682. dx = 1 + (S[6n-2] modulo 399) 5683. dy = 1 + (S[6n-1] modulo 399) 5684. dz = 1 + (S[6n] modulo 399) 5685. 5686. where S,...,S come from the "Lagged Fibonacci Generator": 5687. 5688. For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3] (modulo 1000000) 5689. For 56 ≤ k, S[k] = [S[k-24] + S[k-55]] (modulo 1000000) 5690. 5691. Thus, C has parameters {(7,53,183),(94,369,56)}, C has parameters 5692. {(2383,3563,5079),(42,212,344)}, and so on. 5693. 5694. The combined volume of the first 100 cuboids, C,...,C, is 5695. 723581599. 5696. 5697. What is the combined volume of all 50000 cuboids, C,...,C ? 5698. 5699. 5700. Answer: 76650c9c077929e1ce5a80a1ac81fa96 5701. 5702. 5703. Problem 213 5704. =========== 5705. 5706. 5707. A 30×30 grid of squares contains 900 fleas, initially one flea per square. 5708. When a bell is rung, each flea jumps to an adjacent square at random 5709. (usually 4 possibilities, except for fleas on the edge of the grid or at 5710. the corners). 5711. 5712. What is the expected number of unoccupied squares after 50 rings of the 5713. bell? Give your answer rounded to six decimal places. 5714. 5715. 5716. Answer: f81ee7dd444a3d895a4a446f9d115bf8 5717. 5718. 5719. Problem 214 5720. =========== 5721. 5722. 5723. Let φ be Euler's totient function, i.e. for a natural number n,φ(n) is the 5724. number of k, 1 ≤ k ≤ n, for which gcd(k,n) = 1. 5725. 5726. By iterating φ, each positive integer generates a decreasing chain of 5727. numbers ending in 1. 5728. E.g. if we start with 5 the sequence 5,4,2,1 is generated. 5729. Here is a listing of all chains with length 4: 5730. 5731. 5,4,2,1 5732. 7,6,2,1 5733. 8,4,2,1 5734. 9,6,2,1 5735. 10,4,2,1 5736. 12,4,2,1 5737. 14,6,2,1 5738. 18,6,2,1 5739. 5740. Only two of these chains start with a prime, their sum is 12. 5741. 5742. What is the sum of all primes less than 40000000 which generate a chain of 5743. length 25? 5744. 5745. 5746. Answer: 1cefd865483c03552d5247ffb05685c7 5747. 5748. 5749. Problem 215 5750. =========== 5751. 5752. 5753. Consider the problem of building a wall out of 2×1 and 3×1 bricks 5754. (horizontal×vertical dimensions) such that, for extra strength, the gaps 5755. between horizontally-adjacent bricks never line up in consecutive layers, 5756. i.e. never form a "running crack". 5757. 5758. For example, the following 9×3 wall is not acceptable due to the running 5759. crack shown in red: 5760. 5761. There are eight ways of forming a crack-free 9×3 wall, written W(9,3) = 8. 5762. 5763. Calculate W(32,10). 5764. 5765. 5766. p_215_crackfree.gif 5767. Answer: 60212c9ec4a6cd1d14277c32b6adf2d8 5768. 5769. 5770. Problem 216 5771. =========== 5772. 5773. 5774. Consider numbers t(n) of the form t(n) = 2n^2-1 with n > 1. 5775. The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161. 5776. It turns out that only 49 = 7*7 and 161 = 7*23 are not prime. 5777. For n ≤ 10000 there are 2202 numbers t(n) that are prime. 5778. 5779. How many numbers t(n) are prime for n ≤ 50,000,000 ? 5780. 5781. 5782. Answer: 5437849 5783. 5784. 5785. Problem 217 5786. =========== 5787. 5788. 5789. A positive integer with k (decimal) digits is called balanced if its first 5790. ⌈^k/⌉ digits sum to the same value as its last ⌈^k/⌉ digits, where 5791. ⌈x⌉, pronounced ceiling of x, is the smallest integer ≥ x, thus ⌈π⌉ = 4 5792. and ⌈5⌉ = 5. 5793. 5794. So, for example, all palindromes are balanced, as is 13722. 5795. 5796. Let T(n) be the sum of all balanced numbers less than 10^n. 5797. Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890. 5798. 5799. Find T(47) mod 3^15 5800. 5801. 5802. Answer: 6273134 5803. 5804. 5805. Problem 218 5806. =========== 5807. 5808. 5809. Consider the right angled triangle with sides a=7, b=24 and c=25.The area 5810. of this triangle is 84, which is divisible by the perfect numbers 6 and 5811. 28. 5812. Moreover it is a primitive right angled triangle as gcd(a,b)=1 and 5813. gcd(b,c)=1. 5814. Also c is a perfect square. 5815. 5816. We will call a right angled triangle perfect if 5817. -it is a primitive right angled triangle 5818. -its hypotenuse is a perfect square 5819. 5820. We will call a right angled triangle super-perfect if 5821. -it is a perfect right angled triangle and 5822. -its area is a multiple of the perfect numbers 6 and 28. 5823. 5824. How many perfect right-angled triangles with c≤10^16 exist that are not 5825. super-perfect? 5826. 5827. 5828. Answer: 0 5829. 5830. 5831. Problem 219 5832. =========== 5833. 5834. 5835. Let A and B be bit strings (sequences of 0's and 1's). 5836. If A is equal to the leftmost length(A) bits of B, then A is said to be a 5837. prefix of B. 5838. For example, 00110 is a prefix of 001101001, but not of 00111 or 100110. 5839. 5840. A prefix-free code of size n is a collection of n distinct bit strings 5841. such that no string is a prefix of any other. For example, this is a 5842. prefix-free code of size 6: 5843. 5844. 0000, 0001, 001, 01, 10, 11 5845. 5846. Now suppose that it costs one penny to transmit a '0' bit, but four pence 5847. to transmit a '1'. 5848. Then the total cost of the prefix-free code shown above is 35 pence, which 5849. happens to be the cheapest possible for the skewed pricing scheme in 5850. question. 5851. In short, we write Cost(6) = 35. 5852. 5853. What is Cost(10^9) ? 5854. 5855. 5856. Answer: 578c22ef288b88c60fbcf4541351aff5 5857. 5858. 5859. Problem 220 5860. =========== 5861. 5862. 5863. Let D be the two-letter string "Fa". For n≥1, derive D[n] from D[n-1] 5864. by the string-rewriting rules: 5865. 5866. "a" → "aRbFR" 5867. "b" → "LFaLb" 5868. 5869. Thus, D = "Fa", D = "FaRbFR", D = "FaRbFRRLFaLbFR", and so on. 5870. 5871. These strings can be interpreted as instructions to a computer graphics 5872. program, with "F" meaning "draw forward one unit", "L" meaning "turn left 5873. 90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being 5874. ignored. The initial position of the computer cursor is (0,0), pointing up 5875. towards (0,1). 5876. 5877. Then D[n] is an exotic drawing known as the Heighway Dragon of order n. 5878. For example, D is shown below; counting each "F" as one step, the 5879. highlighted spot at (18,16) is the position reached after 500 steps. 5880. 5881. What is the position of the cursor after 10^12 steps in D ? 5882. Give your answer in the form x,y with no spaces. 5883. 5884. 5885. p_220.gif 5886. Answer: e2018d8efde8ea00319f1adc042f150b 5887. 5888. 5889. Problem 221 5890. =========== 5891. 5892. 5893. We shall call a positive integer A an "Alexandrian integer", if there 5894. exist integers p, q, r such that: 5895. 5896. A = p · q · r and 1 = 1 + 1 + 1 5897. A p q r 5898. 5899. For example, 630 is an Alexandrian integer (p = 5, q = −7, r = −18).In 5900. fact, 630 is the 6^th Alexandrian integer, the first 6 Alexandrian 5901. integers being: 6, 42, 120, 156, 420 and 630. 5902. 5903. Find the 150000^th Alexandrian integer. 5904. 5905. 5906. Answer: cb000c24f653d9c8f78b74123e6515ab 5907. 5908. 5909. Problem 222 5910. =========== 5911. 5912. 5913. What is the length of the shortest pipe, of internal radius 50mm, that can 5914. fully contain 21 balls of radii 30mm, 31mm, ..., 50mm? 5915. 5916. Give your answer in micrometres (10^-6 m) rounded to the nearest integer. 5917. 5918. 5919. Answer: 1590933 5920. 5921. 5922. Problem 223 5923. =========== 5924. 5925. 5926. Let us call an integer sided triangle with sides a ≤ b ≤ c barely acute if 5927. the sides satisfy 5928. a^2 + b^2 = c^2 + 1. 5929. 5930. How many barely acute triangles are there with perimeter ≤ 25,000,000? 5931. 5932. 5933. Answer: 61614848 5934. 5935. 5936. Problem 224 5937. =========== 5938. 5939. 5940. Let us call an integer sided triangle with sides a ≤ b ≤ c barely obtuse 5941. if the sides satisfy 5942. a^2 + b^2 = c^2 - 1. 5943. 5944. How many barely obtuse triangles are there with perimeter ≤ 75,000,000? 5945. 5946. 5947. Answer: 4137330 5948. 5949. 5950. Problem 225 5951. =========== 5952. 5953. 5954. The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ... 5955. is defined by T = T = T = 1 and T[n] = T[n-1] + T[n-2] + T[n-3]. 5956. 5957. It can be shown that 27 does not divide any terms of this sequence. 5958. In fact, 27 is the first odd number with this property. 5959. 5960. Find the 124^th odd number that does not divide any terms of the above 5961. sequence. 5962. 5963. 5964. Answer: 2009 5965. 5966. 5967. Problem 226 5968. =========== 5969. 5970. 5971. The blancmange curve is the set of points (x,y) such that 0 ≤ x ≤ 1 and , 5972. where s(x) = the distance from x to the nearest integer. 5973. 5974. The area under the blancmange curve is equal to ½, shown in pink in the 5975. diagram below. 5976. 5977. blancmange curve 5978. 5979. Let C be the circle with centre (¼,½) and radius ¼, shown in black in the 5980. diagram. 5981. 5982. What area under the blancmange curve is enclosed by C? 5983. Give your answer rounded to eight decimal places in the form 0.abcdefgh 5984. 5985. 5986. Visible links 5987. p_226_formula.gif 5988. p_226_scoop2.gif 5989. Answer: ce6fd32d1d2fb58c4c0c1f7962c39f04 5990. 5991. 5992. Problem 227 5993. =========== 5994. 5995. 5996. "The Chase" is a game played with two dice and an even number of players. 5997. 5998. The players sit around a table; the game begins with two opposite players 5999. having one die each. On each turn, the two players with a die roll it. 6000. If a player rolls a 1, he passes the die to his neighbour on the left; if 6001. he rolls a 6, he passes the die to his neighbour on the right; otherwise, 6002. he keeps the die for the next turn. 6003. The game ends when one player has both dice after they have been rolled 6004. and passed; that player has then lost. 6005. 6006. In a game with 100 players, what is the expected number of turns the game 6007. lasts? 6008. 6009. Give your answer rounded to ten significant digits. 6010. 6011. 6012. Answer: 7b87cd0a96f0f2f12f911cdc66608d95 6013. 6014. 6015. Problem 228 6016. =========== 6017. 6018. 6019. Let S[n] be the regular n-sided polygon – or shape – whose vertices v[k] 6020. (k = 1,2,…,n) have coordinates: 6021. 6022. x[k] = cos( ^2k-1/[n] ×180° ) 6023. y[k] = sin( ^2k-1/[n] ×180° ) 6024. 6025. Each S[n] is to be interpreted as a filled shape consisting of all points 6026. on the perimeter and in the interior. 6027. 6028. The Minkowski sum, S+T, of two shapes S and T is the result of adding 6029. every point in S to every point in T, where point addition is performed 6030. coordinate-wise: (u, v) + (x, y) = (u+x, v+y). 6031. 6032. For example, the sum of S and S is the six-sided shape shown in pink 6033. below: 6034. 6035. picture showing S_3 + S_4 6036. 6037. How many sides does S + S + … + S have? 6038. 6039. 6040. Visible links 6041. p_228.png 6042. Answer: 86226 6043. 6044. 6045. Problem 229 6046. =========== 6047. 6048. 6049. Consider the number 3600. It is very special, because 6050. 6051. 3600 = 48^2 + 36^2 6052. 6053. 3600 = 20^2 + 2×40^2 6054. 6055. 3600 = 30^2 + 3×30^2 6056. 6057. 3600 = 45^2 + 7×15^2 6058. 6059. Similarly, we find that 88201 = 99^2 + 280^2 = 287^2 + 2×54^2 = 283^2 + 6060. 3×52^2 = 197^2 + 7×84^2. 6061. 6062. In 1747, Euler proved which numbers are representable as a sum of two 6063. squares.We are interested in the numbers n which admit representations of 6064. all of the following four types: 6065. 6066. n = a^2 + b^2 6067. 6068. n = a^2 + 2 b^2 6069. 6070. n = a^2 + 3 b^2 6071. 6072. n = a^2 + 7 b^2, 6073. 6074. where the a[k] and b[k] are positive integers. 6075. 6076. There are 75373 such numbers that do not exceed 10^7. 6077. How many such numbers are there that do not exceed 2×10^9? 6078. 6079. 6080. Answer: 11325263 6081. 6082. 6083. Problem 230 6084. =========== 6085. 6086. 6087. For any two strings of digits, A and B, we define F[A,B] to be the 6088. sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of 6089. the previous two. 6090. 6091. Further, we define D[A,B](n) to be the n^th digit in the first term of 6092. F[A,B] that contains at least n digits. 6093. 6094. Example: 6095. 6096. Let A=1415926535, B=8979323846. We wish to find D[A,B](35), say. 6097. 6098. The first few terms of F[A,B] are: 6099. 1415926535 6100. 8979323846 6101. 14159265358979323846 6102. 897932384614159265358979323846 6103. 14159265358979323846897932384614159265358979323846 6104. 6105. Then D[A,B](35) is the 35^th digit in the fifth term, which is 9. 6106. 6107. Now we use for A the first 100 digits of π behind the decimal point: 6108. 6109. 14159265358979323846264338327950288419716939937510 6110. 58209749445923078164062862089986280348253421170679 6111. 6112. and for B the next hundred digits: 6113. 6114. 82148086513282306647093844609550582231725359408128 6115. 48111745028410270193852110555964462294895493038196 . 6116. 6117. Find ∑[n = 0,1,...,17] 10^n× D[A,B]((127+19n)×7^n) . 6118. 6119. 6120. Answer: 040735038021ff4704bbd3a0964369ef 6121. 6122. 6123. Problem 231 6124. =========== 6125. 6126. 6127. The binomial coefficient ^10C = 120. 6128. 120 = 2^3 × 3 × 5 = 2 × 2 × 2 × 3 × 5, and 2 + 2 + 2 + 3 + 5 = 14. 6129. So the sum of the terms in the prime factorisation of ^10C is 14. 6130. 6131. Find the sum of the terms in the prime factorisation of 6132. ^20000000C. 6133. 6134. 6135. Answer: ef8bc4d9a843e71126bd10b5065132a5 6136. 6137. 6138. Problem 232 6139. =========== 6140. 6141. 6142. Two players share an unbiased coin and take it in turns to play "The 6143. Race". On Player 1's turn, he tosses the coin once: if it comes up Heads, 6144. he scores one point; if it comes up Tails, he scores nothing. On Player 6145. 2's turn, she chooses a positive integer T and tosses the coin T times: if 6146. it comes up all Heads, she scores 2^T-1 points; otherwise, she scores 6147. nothing. Player 1 goes first. The winner is the first to 100 or more 6148. points. 6149. 6150. On each turn Player 2 selects the number, T, of coin tosses that maximises 6151. the probability of her winning. 6152. 6153. What is the probability that Player 2 wins? 6154. 6155. Give your answer rounded to eight decimal places in the form 0.abcdefgh . 6156. 6157. 6158. Answer: c8d5b243aa6e6b507725766f7c197a1d 6159. 6160. 6161. Problem 233 6162. =========== 6163. 6164. 6165. Let f(N) be the number of points with integer coordinates that are on a 6166. circle passing through (0,0), (N,0),(0,N), and (N,N). 6167. 6168. It can be shown that f(10000) = 36. 6169. 6170. What is the sum of all positive integers N ≤ 10^11 such that f(N) = 420 ? 6171. 6172. 6173. Answer: 7e80b27798170abb493e3b4671bd82ca 6174. 6175. 6176. Problem 234 6177. =========== 6178. 6179. 6180. For an integer n ≥ 4, we define the lower prime square root of n, denoted 6181. by lps(n), as the largest prime ≤ √n and the upper prime square root of n, 6182. ups(n), as the smallest prime ≥ √n. 6183. 6184. So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. 6185. Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) 6186. divides n, but not both. 6187. 6188. The sum of the semidivisible numbers not exceeding 15 is 30, the numbers 6189. are 8, 10 and 12. 6190. 15 is not semidivisible because it is a multiple of both lps(15) = 3 and 6191. ups(15) = 5. 6192. As a further example, the sum of the 92 semidivisible numbers up to 1000 6193. is 34825. 6194. 6195. What is the sum of all semidivisible numbers not exceeding 999966663333 ? 6196. 6197. 6198. Answer: c24a5d60f8ce5d04dec7466987c84d68 6199. 6200. 6201. Problem 235 6202. =========== 6203. 6204. 6205. Given is the arithmetic-geometric sequence u(k) = (900-3k)r^k-1. 6206. Let s(n) = Σ[k=1...n]u(k). 6207. 6208. Find the value of r for which s(5000) = -600,000,000,000. 6209. 6210. Give your answer rounded to 12 places behind the decimal point. 6211. 6212. 6213. Answer: 41b13508789be1001308e065d4f83ea2 6214. 6215. 6216. Problem 236 6217. =========== 6218. 6219. 6220. Suppliers 'A' and 'B' provided the following numbers of products for the 6221. luxury hamper market: 6222. 6223. Product 'A' 'B' 6224. Beluga Caviar 5248 640 6225. Christmas Cake 1312 1888 6226. Gammon Joint 2624 3776 6227. Vintage Port 5760 3776 6228. Champagne Truffles 3936 5664 6229. 6230. Although the suppliers try very hard to ship their goods in perfect 6231. condition, there is inevitably some spoilage - i.e. products gone bad. 6232. 6233. The suppliers compare their performance using two types of statistic: 6234. 6235. • The five per-product spoilage rates for each supplier are equal to the 6236. number of products gone bad divided by the number of products 6237. supplied, for each of the five products in turn. 6238. • The overall spoilage rate for each supplier is equal to the total 6239. number of products gone bad divided by the total number of products 6240. provided by that supplier. 6241. 6242. To their surprise, the suppliers found that each of the five per-product 6243. spoilage rates was worse (higher) for 'B' than for 'A' by the same factor 6244. (ratio of spoilage rates), m>1; and yet, paradoxically, the overall 6245. spoilage rate was worse for 'A' than for 'B', also by a factor of m. 6246. 6247. There are thirty-five m>1 for which this surprising result could have 6248. occurred, the smallest of which is 1476/1475. 6249. 6250. What's the largest possible value of m? 6251. Give your answer as a fraction reduced to its lowest terms, in the form 6252. u/v. 6253. 6254. 6255. Answer: 6e707fcffc510520d981ae16a29579bb 6256. 6257. 6258. Problem 237 6259. =========== 6260. 6261. 6262. Let T(n) be the number of tours over a 4 × n playing board such that: 6263. 6264. • The tour starts in the top left corner. 6265. • The tour consists of moves that are up, down, left, or right one 6266. square. 6267. • The tour visits each square exactly once. 6268. • The tour ends in the bottom left corner. 6269. 6270. The diagram shows one tour over a 4 × 10 board: 6271. 6272. T(10) is 2329. What is T(10^12) modulo 10^8? 6273. 6274. 6275. p_237.gif 6276. Answer: 15836928 6277. 6278. 6279. Problem 238 6280. =========== 6281. 6282. 6283. Create a sequence of numbers using the "Blum Blum Shub" pseudo-random 6284. number generator: 6285. 6286. s = 14025256 6287. s[n+1] = s[n]^2 mod 20300713 6288. 6289. Concatenate these numbers sss… to create a string w of infinite 6290. length. 6291. Then, w = 14025256741014958470038053646… 6292. 6293. For a positive integer k, if no substring of w exists with a sum of digits 6294. equal to k, p(k) is defined to be zero. If at least one substring of w 6295. exists with a sum of digits equal to k, we define p(k) = z, where z is the 6296. starting position of the earliest such substring. 6297. 6298. For instance: 6299. 6300. The substrings 1, 14, 1402, … 6301. with respective sums of digits equal to 1, 5, 7, … 6302. start at position 1, hence p(1) = p(5) = p(7) = … = 1. 6303. 6304. The substrings 4, 402, 4025, … 6305. with respective sums of digits equal to 4, 6, 11, … 6306. start at position 2, hence p(4) = p(6) = p(11) = … = 2. 6307. 6308. The substrings 02, 0252, … 6309. with respective sums of digits equal to 2, 9, … 6310. start at position 3, hence p(2) = p(9) = … = 3. 6311. 6312. Note that substring 025 starting at position 3, has a sum of digits equal 6313. to 7, but there was an earlier substring (starting at position 1) with a 6314. sum of digits equal to 7, so p(7) = 1, not 3. 6315. 6316. We can verify that, for 0 < k ≤ 10^3, ∑ p(k) = 4742. 6317. 6318. Find ∑ p(k), for 0 < k ≤ 2·10^15. 6319. 6320. 6321. Answer: 424ed6613a372ccb9a90dddb8961ca16 6322. 6323. 6324. Problem 239 6325. =========== 6326. 6327. 6328. A set of disks numbered 1 through 100 are placed in a line in random 6329. order. 6330. 6331. What is the probability that we have a partial derangement such that 6332. exactly 22 prime number discs are found away from their natural positions? 6333. (Any number of non-prime disks may also be found in or out of their 6334. natural positions.) 6335. 6336. Give your answer rounded to 12 places behind the decimal point in the form 6337. 0.abcdefghijkl. 6338. 6339. 6340. Answer: 451fd2b8c19fbfec650a5c4699f6ef6e 6341. 6342. 6343. Problem 240 6344. =========== 6345. 6346. 6347. There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can 6348. be rolled so that the top three sum to 15. Some examples are: 6349. 6350. D,D,D,D,D = 4,3,6,3,5 6351. D,D,D,D,D = 4,3,3,5,6 6352. D,D,D,D,D = 3,3,3,6,6 6353. D,D,D,D,D = 6,6,3,3,3 6354. 6355. In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be 6356. rolled so that the top ten sum to 70? 6357. 6358. 6359. Answer: cb31a3106db3876e77cd160664cd683e 6360. 6361. 6362. Problem 241 6363. =========== 6364. 6365. 6366. For a positive integer n, let σ(n) be the sum of all divisors of n, so 6367. e.g. σ(6) = 1 + 2 + 3 + 6 = 12. 6368. 6369. A perfect number, as you probably know, is a number with σ(n) = 2n. 6370. 6371. Let us define the perfection quotient of a positive integer p(n) = σ(n) . 6372. as n 6373. 6374. Find the sum of all positive integers n ≤ 10^18 for which p(n) has the 6375. form k + ^1⁄, where k is an integer. 6376. 6377. 6378. Answer: 556bfef2cacd1eff8af9126c5c13dcbc 6379. 6380. 6381. Problem 242 6382. =========== 6383. 6384. 6385. Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element 6386. subsets with an odd sum of elements. For example, f(5,3) = 4, since the 6387. set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements, 6388. i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}. 6389. 6390. When all three values n, k and f(n,k) are odd, we say that they make 6391. an odd-triplet [n,k,f(n,k)]. 6392. 6393. There are exactly five odd-triplets with n ≤ 10, namely: 6394. [1,1,f(1,1) = 1], [5,1,f(5,1) = 3], [5,5,f(5,5) = 1], [9,1,f(9,1) = 5] and 6395. [9,9,f(9,9) = 1]. 6396. 6397. How many odd-triplets are there with n ≤ 10^12 ? 6398. 6399. 6400. Answer: ba73cb75365ddca8f94a23e3fedfb6de 6401. 6402. 6403. Problem 243 6404. =========== 6405. 6406. 6407. A positive fraction whose numerator is less than its denominator is called 6408. a proper fraction. 6409. For any denominator, d, there will be d−1 proper fractions; for example, 6410. with d = 12: 6411. 1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , 6412. 8/12 , 9/12 , 10/12 , 11/12 . 6413. 6414. We shall call a fraction that cannot be cancelled down a resilient 6415. fraction. 6416. Furthermore we shall define the resilience of a denominator, R(d), to be 6417. the ratio of its proper fractions that are resilient; for example, R(12) = 6418. 4/11 . 6419. In fact, d = 12 is the smallest denominator having a resilience R(d) < 6420. 4/10 . 6421. 6422. Find the smallest denominator d, having a resilience R(d) < 15499/94744 6423. . 6424. 6425. 6426. Answer: 531721a10786c5c2a444b474fcf039f9 6427. 6428. 6429. Problem 244 6430. =========== 6431. 6432. 6433. You probably know the game Fifteen Puzzle. Here, instead of numbered 6434. tiles, we have seven red tiles and eight blue tiles. 6435. 6436. A move is denoted by the uppercase initial of the direction (Left, Right, 6437. Up, Down) in which the tile is slid, e.g. starting from configuration (S), 6438. by the sequence LULUR we reach the configuration (E): 6439. 6440. (S) , (E) 6441. 6442. For each path, its checksum is calculated by (pseudocode): 6443. checksum = 0 6444. checksum = (checksum × 243 + m) mod 100 000 007 6445. checksum = (checksum × 243 + m) mod 100 000 007 6446. … 6447. checksum = (checksum × 243 + m[n]) mod 100 000 007 6448. where m[k] is the ASCII value of the k^th letter in the move sequence and 6449. the ASCII values for the moves are: 6450. 6451. ┌──────┬─────┐ 6452. │L │76 │ 6453. ├──────┼─────┤ 6454. │R │82 │ 6455. ├──────┼─────┤ 6456. │U │85 │ 6457. ├──────┼─────┤ 6458. │D │68 │ 6459. └──────┴─────┘ 6460. 6461. For the sequence LULUR given above, the checksum would be 19761398. 6462. 6463. Now, starting from configuration (S),find all shortest ways to reach 6464. configuration (T). 6465. 6466. (S) , (T) 6467. 6468. What is the sum of all checksums for the paths having the minimal length? 6469. 6470. 6471. p_244_start.gif 6472. p_244_example.gif 6473. p_244_start.gif 6474. p_244_target.gif 6475. Answer: 96356848 6476. 6477. 6478. Problem 245 6479. =========== 6480. 6481. 6482. We shall call a fraction that cannot be cancelled down a resilient 6483. fraction. 6484. Furthermore we shall define the resilience of a denominator, R(d), to be 6485. the ratio of its proper fractions that are resilient; for example, R(12) = 6486. ^4⁄. 6487. 6488. The resilience of a number d > 1 is φ(d) , where φ is Euler's totient 6489. then d - 1 function. 6490. 6491. We further define the coresilience of a number n > 1 as C(n) = n - φ(n) . 6492. n - 1 6493. 6494. The coresilience of a prime p is C(p) = 1 . 6495. p - 1 6496. 6497. Find the sum of all composite integers 1 < n ≤ 2×10^11, for which C(n) is 6498. a unit fraction. 6499. 6500. 6501. Answer: 0ebeb502fb0bd7157609835d27c266bc 6502. 6503. 6504. Problem 246 6505. =========== 6506. 6507. 6508. A definition for an ellipse is: 6509. Given a circle c with centre M and radius r and a point G such that 6510. d(G,M)<r, the locus of the points that are equidistant from c and G form 6511. an ellipse. 6512. 6513. The construction of the points of the ellipse is shown below. 6514. 6515. Given are the points M(-2000,1500) and G(8000,1500). 6516. Given is also the circle c with centre M and radius 15000. 6517. The locus of the points that are equidistant from G and c form an ellipse 6518. e. 6519. From a point P outside e the two tangents t and t to the ellipse are 6520. drawn. 6521. Let the points where t and t touch the ellipse be R and S. 6522. 6523. For how many lattice points P is angle RPS greater than 45 degrees? 6524. 6525. 6526. p_246_anim.gif 6527. p_246_ellipse.gif 6528. Answer: 94c521ffeb906391d161b66fec433827 6529. 6530. 6531. Problem 247 6532. =========== 6533. 6534. 6535. Consider the region constrained by 1 ≤ x and 0 ≤ y ≤ ^1/[x]. 6536. 6537. Let S be the largest square that can fit under the curve. 6538. Let S be the largest square that fits in the remaining area, and so on. 6539. Let the index of S[n] be the pair (left, below) indicating the number of 6540. squares to the left of S[n] and the number of squares below S[n]. 6541. 6542. The diagram shows some such squares labelled by number. 6543. S has one square to its left and none below, so the index of S is 6544. (1,0). 6545. It can be seen that the index of S is (1,1) as is the index of S. 6546. 50 is the largest n for which the index of S[n] is (1,1). 6547. 6548. What is the largest n for which the index of S[n] is (3,3)? 6549. 6550. 6551. p_247_hypersquares.gif 6552. Answer: 782252 6553. 6554. 6555. Problem 248 6556. =========== 6557. 6558. 6559. The first number n for which φ(n)=13! is 6227180929. 6560. 6561. Find the 150,000^th such number. 6562. 6563. 6564. Answer: b69a3ba674f6c7c5f2ce244f9e9cc873 6565. 6566. 6567. Problem 249 6568. =========== 6569. 6570. 6571. Let S = {2, 3, 5, ..., 4999} be the set of prime numbers less than 5000. 6572. 6573. Find the number of subsets of S, the sum of whose elements is a prime 6574. number. 6575. Enter the rightmost 16 digits as your answer. 6576. 6577. 6578. Answer: a470ee3ca52f2b68d7034e48b39b8b26 6579. 6580. 6581. Problem 250 6582. =========== 6583. 6584. 6585. Find the number of non-empty subsets of {1^1, 2^2, 3^3,..., 6586. 250250^250250}, the sum of whose elements is divisible by 250. Enter the 6587. rightmost 16 digits as your answer. 6588. 6589. 6590. Answer: 4a5614f3700956273fe0d271f921d5f4 6591. 6592. 6593. Problem 251 6594. =========== 6595. 6596. 6597. A triplet of positive integers (a,b,c) is called a Cardano Triplet if it 6598. satisfies the condition: 6599. 6600. For example, (2,1,5) is a Cardano Triplet. 6601. 6602. There exist 149 Cardano Triplets for which a+b+c ≤ 1000. 6603. 6604. Find how many Cardano Triplets exist such that a+b+c ≤ 110,000,000. 6605. 6606. 6607. p_251_cardano.gif 6608. Answer: 18946051 6609. 6610. 6611. Problem 252 6612. =========== 6613. 6614. 6615. Given a set of points on a plane, we define a convex hole to be a convex 6616. polygon having as vertices any of the given points and not containing any 6617. of the given points in its interior (in addition to the vertices, other 6618. given points may lie on the perimeter of the polygon). 6619. 6620. As an example, the image below shows a set of twenty points and a few such 6621. convex holes. The convex hole shown as a red heptagon has an area equal to 6622. 1049694.5 square units, which is the highest possible area for a convex 6623. hole on the given set of points. 6624. 6625. For our example, we used the first 20 points (T[2k−1], T[2k]), for 6626. k = 1,2,…,20, produced with the pseudo-random number generator: 6627. 6628. S =[ ] 290797[ ] 6629. S[n+1] =[ ] S[n]^2 mod 50515093 6630. T[n] =[ ] ( S[n] mod 2000 ) − 1000^ 6631. 6632. i.e. (527, 144), (−488, 732), (−454, −947), … 6633. 6634. What is the maximum area for a convex hole on the set containing the first 6635. 500 points in the pseudo-random sequence? 6636. Specify your answer including one digit after the decimal point. 6637. 6638. 6639. p_252_convexhole.gif 6640. Answer: 53b1ced82e1b588d756750c4d2f77e0d 6641. 6642. 6643. Problem 253 6644. =========== 6645. 6646. 6647. A small child has a “number caterpillar” consisting of forty jigsaw 6648. pieces, each with one number on it, which, when connected together in a 6649. line, reveal the numbers 1 to 40 in order. 6650. 6651. Every night, the child's father has to pick up the pieces of the 6652. caterpillar that have been scattered across the play room. He picks up the 6653. pieces at random and places them in the correct order. 6654. As the caterpillar is built up in this way, it forms distinct segments 6655. that gradually merge together. 6656. The number of segments starts at zero (no pieces placed), generally 6657. increases up to about eleven or twelve, then tends to drop again before 6658. finishing at a single segment (all pieces placed). 6659. 6660. For example: 6661. 6662. ┌────────────┬───────────────┐ 6663. │Piece Placed│Segments So Far│ 6664. ├────────────┼───────────────┤ 6665. │ 12 │ 1 │ 6666. ├────────────┼───────────────┤ 6667. │ 4 │ 2 │ 6668. ├────────────┼───────────────┤ 6669. │ 29 │ 3 │ 6670. ├────────────┼───────────────┤ 6671. │ 6 │ 4 │ 6672. ├────────────┼───────────────┤ 6673. │ 34 │ 5 │ 6674. ├────────────┼───────────────┤ 6675. │ 5 │ 4 │ 6676. ├────────────┼───────────────┤ 6677. │ 35 │ 4 │ 6678. ├────────────┼───────────────┤ 6679. │ … │ … │ 6680. └────────────┴───────────────┘ 6681. 6682. Let M be the maximum number of segments encountered during a random 6683. tidy-up of the caterpillar. 6684. For a caterpillar of ten pieces, the number of possibilities for each M is 6685. 6686. ┌────────┬─────────────┐ 6687. │ M │Possibilities│ 6688. ├────────┼─────────────┤ 6689. │ 1 │ 512 │ 6690. ├────────┼─────────────┤ 6691. │ 2 │ 250912 │ 6692. ├────────┼─────────────┤ 6693. │ 3 │1815264 │ 6694. ├────────┼─────────────┤ 6695. │ 4 │1418112 │ 6696. ├────────┼─────────────┤ 6697. │ 5 │ 144000 │ 6698. └────────┴─────────────┘ 6699. 6700. so the most likely value of M is 3 and the average value is 6701. ^385643⁄ = 3.400732, rounded to six decimal places. 6702. 6703. The most likely value of M for a forty-piece caterpillar is 11; but what 6704. is the average value of M? 6705. 6706. Give your answer rounded to six decimal places. 6707. 6708. 6709. Answer: 228de0a37019fd7c7051029f3d126422 6710. 6711. 6712. Problem 254 6713. =========== 6714. 6715. 6716. Define f(n) as the sum of the factorials of the digits of n. For example, 6717. f(342) = 3! + 4! + 2! = 32. 6718. 6719. Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5. 6720. 6721. Define g(i) to be the smallest positive integer n such that sf(n) = i. 6722. Though sf(342) is 5, sf(25) is also 5, and it can be verified that g(5) is 6723. 25. 6724. 6725. Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7. 6726. 6727. Further, it can be verified that g(20) is 267 and ∑ sg(i) for 1 ≤ i ≤ 20 6728. is 156. 6729. 6730. What is ∑ sg(i) for 1 ≤ i ≤ 150? 6731. 6732. 6733. Answer: 936014adf2de65d41979ad900325e485 6734. 6735. 6736. Problem 255 6737. =========== 6738. 6739. 6740. We define the rounded-square-root of a positive integer n as the square 6741. root of n rounded to the nearest integer. 6742. 6743. The following procedure (essentially Heron's method adapted to integer 6744. arithmetic) finds the rounded-square-root of n: 6745. 6746. Let d be the number of digits of the number n. 6747. If d is odd, set x = 2×10^(d-1)⁄2. 6748. If d is even, set x = 7×10^(d-2)⁄2. 6749. Repeat: 6750. 6751. until x[k+1] = x[k]. 6752. 6753. As an example, let us find the rounded-square-root of n = 4321. 6754. n has 4 digits, so x = 7×10^(4-2)⁄2 = 70. 6755. Since x = x, we stop here. 6756. So, after just two iterations, we have found that the rounded-square-root 6757. of 4321 is 66 (the actual square root is 65.7343137…). 6758. 6759. The number of iterations required when using this method is surprisingly 6760. low. 6761. For example, we can find the rounded-square-root of a 5-digit integer 6762. (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the 6763. average value was rounded to 10 decimal places). 6764. 6765. Using the procedure described above, what is the average number of 6766. iterations required to find the rounded-square-root of a 14-digit number 6767. (10^13 ≤ n < 10^14)? 6768. Give your answer rounded to 10 decimal places. 6769. 6770. Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling 6771. function respectively. 6772. 6773. 6774. p_255_Heron.gif 6775. p_255_Example.gif 6776. Answer: 12be028b156b49faa1137febda940ab5 6777. 6778. 6779. Problem 256 6780. =========== 6781. 6782. 6783. Tatami are rectangular mats, used to completely cover the floor of a room, 6784. without overlap. 6785. 6786. Assuming that the only type of available tatami has dimensions 1×2, there 6787. are obviously some limitations for the shape and size of the rooms that 6788. can be covered. 6789. 6790. For this problem, we consider only rectangular rooms with integer 6791. dimensions a, b and even size s = a·b. 6792. We use the term 'size' to denote the floor surface area of the room, and — 6793. without loss of generality — we add the condition a ≤ b. 6794. 6795. There is one rule to follow when laying out tatami: there must be no 6796. points where corners of four different mats meet. 6797. For example, consider the two arrangements below for a 4×4 room: 6798. 6799. The arrangement on the left is acceptable, whereas the one on the right is 6800. not: a red "X" in the middle, marks the point where four tatami meet. 6801. 6802. Because of this rule, certain even-sized rooms cannot be covered with 6803. tatami: we call them tatami-free rooms. 6804. Further, we define T(s) as the number of tatami-free rooms of size s. 6805. 6806. The smallest tatami-free room has size s = 70 and dimensions 7×10. 6807. All the other rooms of size s = 70 can be covered with tatami; they are: 6808. 1×70, 2×35 and 5×14. 6809. Hence, T(70) = 1. 6810. 6811. Similarly, we can verify that T(1320) = 5 because there are exactly 5 6812. tatami-free rooms of size s = 1320: 6813. 20×66, 22×60, 24×55, 30×44 and 33×40. 6814. In fact, s = 1320 is the smallest room-size s for which T(s) = 5. 6815. 6816. Find the smallest room-size s for which T(s) = 200. 6817. 6818. 6819. p_256_tatami3.gif 6820. Answer: 85765680 6821. 6822. 6823. Problem 257 6824. =========== 6825. 6826. 6827. Given is an integer sided triangle ABC with sides a ≤ b ≤ c. (AB = c, BC = 6828. a and AC = b). 6829. The angular bisectors of the triangle intersect the sides at points E, F 6830. and G (see picture below). 6831. 6832. The segments EF, EG and FG partition the triangle ABC into four smaller 6833. triangles: AEG, BFE, CGF and EFG. 6834. It can be proven that for each of these four triangles the ratio 6835. area(ABC)/area(subtriangle) is rational. 6836. However, there exist triangles for which some or all of these ratios are 6837. integral. 6838. 6839. How many triangles ABC with perimeter≤100,000,000 exist so that the ratio 6840. area(ABC)/area(AEG) is integral? 6841. 6842. 6843. p_257_bisector.gif 6844. Answer: 139012411 6845. 6846. 6847. Problem 258 6848. =========== 6849. 6850. 6851. A sequence is defined as: 6852. 6853. • g[k] = 1, for 0 ≤ k ≤ 1999 6854. • g[k] = g[k-2000] + g[k-1999], for k ≥ 2000. 6855. 6856. Find g[k] mod 20092010 for k = 10^18. 6857. 6858. 6859. Answer: 12747994 6860. 6861. 6862. Problem 259 6863. =========== 6864. 6865. 6866. A positive integer will be called reachable if it can result from an 6867. arithmetic expression obeying the following rules: 6868. 6869. • Uses the digits 1 through 9, in that order and exactly once each. 6870. • Any successive digits can be concatenated (for example, using the 6871. digits 2, 3 and 4 we obtain the number 234). 6872. • Only the four usual binary arithmetic operations (addition, 6873. subtraction, multiplication and division) are allowed. 6874. • Each operation can be used any number of times, or not at all. 6875. • Unary minus is not allowed. 6876. • Any number of (possibly nested) parentheses may be used to define the 6877. order of operations. 6878. 6879. For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42. 6880. 6881. What is the sum of all positive reachable integers? 6882. 6883. 6884. Answer: 771828a57c269d873335c9091af78f76 6885. 6886. 6887. Problem 260 6888. =========== 6889. 6890. 6891. A game is played with three piles of stones and two players. 6892. At her turn, a player removes one or more stones from the piles. However, 6893. if she takes stones from more than one pile, she must remove the same 6894. number of stones from each of the selected piles. 6895. 6896. In other words, the player chooses some N>0 and removes: 6897. 6898. • N stones from any single pile; or 6899. • N stones from each of any two piles (2N total); or 6900. • N stones from each of the three piles (3N total). 6901. 6902. The player taking the last stone(s) wins the game. 6903. 6904. A winning configuration is one where the first player can force a win. 6905. For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations 6906. because the first player can immediately remove all stones. 6907. 6908. A losing configuration is one where the second player can force a win, no 6909. matter what the first player does. 6910. For example, (0,1,2) and (1,3,3) are losing configurations: any legal move 6911. leaves a winning configuration for the second player. 6912. 6913. Consider all losing configurations (x[i],y[i],z[i]) where x[i] ≤ y[i] ≤ 6914. z[i] ≤ 100. 6915. We can verify that Σ(x[i]+y[i]+z[i]) = 173895 for these. 6916. 6917. Find Σ(x[i]+y[i]+z[i]) where (x[i],y[i],z[i]) ranges over the losing 6918. configurations 6919. with x[i] ≤ y[i] ≤ z[i] ≤ 1000. 6920. 6921. 6922. Answer: 167542057 6923. 6924. 6925. Problem 261 6926. =========== 6927. 6928. 6929. Let us call a positive integer k a square-pivot, if there is a pair of 6930. integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive 6931. squares up to k equals the sum of the m consecutive squares from (n+1) on: 6932. 6933. (k-m)^2 + ... + k^2 = (n+1)^2 + ... + (n+m)^2. 6934. 6935. Some small square-pivots are 6936. 6937. • 4: 3^2 + 4^2 = 5^2 6938. • 21: 20^2 + 21^2 = 29^2 6939. • 24: 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2 6940. • 110: 108^2 + 109^2 + 110^2 = 133^2 + 134^2 6941. 6942. Find the sum of all distinct square-pivots ≤ 10^10. 6943. 6944. 6945. Answer: d45ddf64010ed143228a6a6b84837de9 6946. 6947. 6948. Problem 262 6949. =========== 6950. 6951. 6952. The following equation represents the continuous topography of a 6953. mountainous region, giving the elevation h at any point (x,y): 6954. 6955. A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving 6956. the area given by 0 ≤ x, y ≤ 1600. 6957. 6958. Because of the intervening mountains, it first rises straight up to a 6959. point A', having elevation f. Then, while remaining at the same elevation 6960. f, it flies around any obstacles until it arrives at a point B' directly 6961. above B. 6962. 6963. First, determine f[min] which is the minimum constant elevation allowing 6964. such a trip from A to B, while remaining in the specified area. 6965. Then, find the length of the shortest path between A' and B', while flying 6966. at that constant elevation f[min]. 6967. 6968. Give that length as your answer, rounded to three decimal places. 6969. 6970. Note: For convenience, the elevation function shown above is repeated 6971. below, in a form suitable for most programming languages: 6972. h=( 5000-0.005*(x*x+y*y+x*y)+12.5*(x+y) ) * exp( 6973. -abs(0.000001*(x*x+y*y)-0.0015*(x+y)+0.7) ) 6974. 6975. 6976. p_262_formula1.gif 6977. Answer: a5921e175a44d31e7f82f7f9a61a36af 6978. 6979. 6980. Problem 263 6981. =========== 6982. 6983. 6984. Consider the number 6. The divisors of 6 are: 1,2,3 and 6. 6985. Every number from 1 up to and including 6 can be written as a sum of 6986. distinct divisors of 6: 6987. 1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6. 6988. A number n is called a practical number if every number from 1 up to and 6989. including n can be expressed as a sum of distinct divisors of n. 6990. 6991. A pair of consecutive prime numbers with a difference of six is called a 6992. sexy pair (since "sex" is the Latin word for "six"). The first sexy pair 6993. is (23, 29). 6994. 6995. We may occasionally find a triple-pair, which means three consecutive sexy 6996. prime pairs, such that the second member of each pair is the first member 6997. of the next pair. 6998. 6999. We shall call a number n such that : 7000. 7001. • (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and 7002. • the numbers n-8, n-4, n, n+4 and n+8 are all practical, 7003. 7004. an engineers’ paradise. 7005. 7006. Find the sum of the first four engineers’ paradises. 7007. 7008. 7009. Answer: 8fe3eb7196c69a080740e076cff9b4a1 7010. 7011. 7012. Problem 264 7013. =========== 7014. 7015. 7016. Consider all the triangles having: 7017. 7018. • All their vertices on lattice points. 7019. • Circumcentre at the origin O. 7020. • Orthocentre at the point H(5, 0). 7021. 7022. There are nine such triangles having a perimeter ≤ 50. 7023. Listed and shown in ascending order of their perimeter, they are: 7024. 7025. A(-4, 3), B(5, 0), C(4, -3) 7026. A(4, 3), B(5, 0), C(-4, -3) 7027. A(-3, 4), B(5, 0), C(3, -4) 7028. 7029. A(3, 4), B(5, 0), C(-3, -4) 7030. A(0, 5), B(5, 0), C(0, -5) 7031. A(1, 8), B(8, -1), C(-4, -7) 7032. 7033. A(8, 1), B(1, -8), C(-4, 7) 7034. A(2, 9), B(9, -2), C(-6, -7) 7035. A(9, 2), B(2, -9), C(-6, 7) 7036. 7037. The sum of their perimeters, rounded to four decimal places, is 291.0089. 7038. 7039. Find all such triangles with a perimeter ≤ 10^5. 7040. Enter as your answer the sum of their perimeters rounded to four decimal 7041. places. 7042. 7043. 7044. p_264_TriangleCentres.gif 7045. Answer: 287514a045a38be0a75a1786694c77ee 7046. 7047. 7048. Problem 265 7049. =========== 7050. 7051. 7052. 2^N binary digits can be placed in a circle so that all the N-digit 7053. clockwise subsequences are distinct. 7054. 7055. For N=3, two such circular arrangements are possible, ignoring rotations: 7056. 7057. For the first arrangement, the 3-digit subsequences, in clockwise order, 7058. are: 7059. 000, 001, 010, 101, 011, 111, 110 and 100. 7060. 7061. Each circular arrangement can be encoded as a number by concatenating the 7062. binary digits starting with the subsequence of all zeros as the most 7063. significant bits and proceeding clockwise. The two arrangements for N=3 7064. are thus represented as 23 and 29: 7065. 00010111  = 23 7066. 00011101  = 29 7067. 7068. Calling S(N) the sum of the unique numeric representations, we can see 7069. that S(3) = 23 + 29 = 52. 7070. 7071. Find S(5). 7072. 7073. 7074. p_265_BinaryCircles.gif 7075. Answer: c25cebbc8dce4bdcf96cb395a11afcc3 7076. 7077. 7078. Problem 266 7079. =========== 7080. 7081. 7082. The divisors of 12 are: 1,2,3,4,6 and 12. 7083. The largest divisor of 12 that does not exceed the square root of 12 is 3. 7084. We shall call the largest divisor of an integer n that does not exceed the 7085. square root of n the pseudo square root (PSR) of n. 7086. It can be seen that PSR(3102)=47. 7087. 7088. Let p be the product of the primes below 190. 7089. Find PSR(p) mod 10^16. 7090. 7091. 7092. Answer: 32da1d501e539ab509f104e2db68d57a 7093. 7094. 7095. Problem 267 7096. =========== 7097. 7098. 7099. You are given a unique investment opportunity. 7100. 7101. Starting with £1 of capital, you can choose a fixed proportion, f, of your 7102. capital to bet on a fair coin toss repeatedly for 1000 tosses. 7103. 7104. Your return is double your bet for heads and you lose your bet for tails. 7105. 7106. For example, if f = 1/4, for the first toss you bet £0.25, and if heads 7107. comes up you win £0.5 and so then have £1.5. You then bet £0.375 and if 7108. the second toss is tails, you have £1.125. 7109. 7110. Choosing f to maximize your chances of having at least £1,000,000,000 7111. after 1,000 flips, what is the chance that you become a billionaire? 7112. 7113. All computations are assumed to be exact (no rounding), but give your 7114. answer rounded to 12 digits behind the decimal point in the form 7115. 0.abcdefghijkl. 7116. 7117. 7118. Answer: b8dd3306c2c64eacb0ac36b414892edb 7119. 7120. 7121. Problem 268 7122. =========== 7123. 7124. 7125. It can be verified that there are 23 positive integers less than 1000 that 7126. are divisible by at least four distinct primes less than 100. 7127. 7128. Find how many positive integers less than 10^16 are divisible by at least 7129. four distinct primes less than 100. 7130. 7131. 7132. Answer: 6f84b20c10311cb24a824416a3c3e0a4 7133. 7134. 7135. Problem 269 7136. =========== 7137. 7138. 7139. A root or zero of a polynomial P(x) is a solution to the equation P(x) = 7140. 0. 7141. Define P[n] as the polynomial whose coefficients are the digits of n. 7142. For example, P(x) = 5x^3 + 7x^2 + 3. 7143. 7144. We can see that: 7145. 7146. • P[n](0) is the last digit of n, 7147. • P[n](1) is the sum of the digits of n, 7148. • P[n](10) is n itself. 7149. 7150. Define Z(k) as the number of positive integers, n, not exceeding k for 7151. which the polynomial P[n] has at least one integer root. 7152. 7153. It can be verified that Z(100 000) is 14696. 7154. 7155. What is Z(10^16)? 7156. 7157. 7158. Answer: f7ba868cb52a9b9c7e58b1b92e230be8 7159. 7160. 7161. Problem 270 7162. =========== 7163. 7164. 7165. A square piece of paper with integer dimensions N×N is placed with a 7166. corner at the origin and two of its sides along the x- and y-axes. Then, 7167. we cut it up respecting the following rules: 7168. 7169. • We only make straight cuts between two points lying on different sides 7170. of the square, and having integer coordinates. 7171. • Two cuts cannot cross, but several cuts can meet at the same border 7172. point. 7173. • Proceed until no more legal cuts can be made. 7174. 7175. Counting any reflections or rotations as distinct, we call C(N) the number 7176. of ways to cut an N×N square. For example, C(1) = 2 and C(2) = 30 (shown 7177. below). 7178. 7179. What is C(30) mod 10^8 ? 7180. 7181. 7182. p_270_CutSquare.gif 7183. Answer: 82282080 7184. 7185. 7186. Problem 271 7187. =========== 7188. 7189. 7190. For a positive number n, define S(n) as the sum of the integers x, for 7191. which 1<x<n and 7192. x^3≡1 mod n. 7193. 7194. When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 7195. 74, 79, 81. 7196. Thus, S(91)=9+16+22+29+53+74+79+81=363. 7197. 7198. Find S(13082761331670030). 7199. 7200. 7201. Answer: c4157aab542bd0dfa465c890e1286cc5 7202. 7203. 7204. Problem 272 7205. =========== 7206. 7207. 7208. For a positive number n, define C(n) as the number of the integers x, for 7209. which 1<x<n and 7210. x^3≡1 mod n. 7211. 7212. When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 7213. 74, 79, 81. 7214. Thus, C(91)=8. 7215. 7216. Find the sum of the positive numbers n≤10^11 for which C(n)=242. 7217. 7218. 7219. Answer: d84d2020055b3e8867dc359e739e0312 7220. 7221. 7222. Problem 273 7223. =========== 7224. 7225. 7226. Consider equations of the form: a^2 + b^2 = N, 0 ≤ a ≤ b, a, b and N 7227. integer. 7228. 7229. For N=65 there are two solutions: 7230. 7231. a=1, b=8 and a=4, b=7. 7232. 7233. We call S(N) the sum of the values of a of all solutions of a^2 + b^2 = N, 7234. 0 ≤ a ≤ b, a, b and N integer. 7235. 7236. Thus S(65) = 1 + 4 = 5. 7237. 7238. Find ∑S(N), for all squarefree N only divisible by primes of the form 4k+1 7239. with 4k+1 < 150. 7240. 7241. 7242. Answer: 2b03731e58e9d60e559ee5fdce4f0d14 7243. 7244. 7245. Problem 274 7246. =========== 7247. 7248. 7249. For each integer p > 1 coprime to 10 there is a positive divisibility 7250. multiplier m < p which preserves divisibility by p for the following 7251. function on any positive integer, n: 7252. 7253. f(n) = (all but the last digit of n) + (the last digit of n) * m 7254. 7255. That is, if m is the divisibility multiplier for p, then f(n) is divisible 7256. by p if and only if n is divisible by p. 7257. 7258. (When n is much larger than p, f(n) will be less than n and repeated 7259. application of f provides a multiplicative divisibility test for p.) 7260. 7261. For example, the divisibility multiplier for 113 is 34. 7262. 7263. f(76275) = 7627 + 5 * 34 = 7797 : 76275 and 7797 are both divisible by 113 7264. f(12345) = 1234 + 5 * 34 = 1404 : 12345 and 1404 are both not divisible by 7265. 113 7266. 7267. The sum of the divisibility multipliers for the primes that are coprime to 7268. 10 and less than 1000 is 39517. What is the sum of the divisibility 7269. multipliers for the primes that are coprime to 10 and less than 10^7? 7270. 7271. 7272. Answer: ffd68ca67b9c3ea2653d375051e70288 7273. 7274. 7275. Problem 275 7276. =========== 7277. 7278. 7279. Let us define a balanced sculpture of order n as follows: 7280. 7281. • A polyomino made up of n+1 tiles known as the blocks (n tiles) 7282. and the plinth (remaining tile); 7283. • the plinth has its centre at position (x = 0, y = 0); 7284. • the blocks have y-coordinates greater than zero (so the plinth is the 7285. unique lowest tile); 7286. • the centre of mass of all the blocks, combined, has x-coordinate equal 7287. to zero. 7288. 7289. When counting the sculptures, any arrangements which are simply 7290. reflections about the y-axis, are not counted as distinct. For example, 7291. the 18 balanced sculptures of order 6 are shown below; note that each pair 7292. of mirror images (about the y-axis) is counted as one sculpture: 7293. 7294. There are 964 balanced sculptures of order 10 and 360505 of order 15. 7295. How many balanced sculptures are there of order 18? 7296. 7297. 7298. p_275_sculptures2.gif 7299. Answer: 15030564 7300. 7301. 7302. Problem 276 7303. =========== 7304. 7305. 7306. Consider the triangles with integer sides a, b and c with a ≤ b ≤ c. 7307. An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. 7308. How many primitive integer sided triangles exist with a perimeter not 7309. exceeding 10 000 000? 7310. 7311. 7312. Answer: 29ae64e74ebfdf459dac56786e95c5d5 7313. 7314. 7315. Problem 277 7316. =========== 7317. 7318. 7319. A modified Collatz sequence of integers is obtained from a starting value 7320. a in the following way: 7321. 7322. a[n+1] = a[n]/3 if a[n] is divisible by 3. We shall denote this as a large 7323. downward step, "D". 7324. 7325. a[n+1] = (4a[n] + 2)/3 if a[n] divided by 3 gives a remainder of 1. We 7326. shall denote this as an upward step, "U". 7327. 7328. a[n+1] = (2a[n] - 1)/3 if a[n] divided by 3 gives a remainder of 2. We 7329. shall denote this as a small downward step, "d". 7330. 7331. The sequence terminates when some a[n] = 1. 7332. 7333. Given any integer, we can list out the sequence of steps. 7334. For instance if a=231, then the sequence 7335. {a[n]}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps 7336. "DdDddUUdDD". 7337. 7338. Of course, there are other sequences that begin with that same sequence 7339. "DdDddUUdDD....". 7340. For instance, if a=1004064, then the sequence is 7341. DdDddUUdDDDdUDUUUdDdUUDDDUdDD. 7342. In fact, 1004064 is the smallest possible a > 10^6 that begins with the 7343. sequence DdDddUUdDD. 7344. 7345. What is the smallest a > 10^15 that begins with the sequence 7346. "UDDDUdddDDUDDddDdDddDDUDDdUUDd"? 7347. 7348. 7349. Answer: 9508afff135320c18d82c93a8b70cd11 7350. 7351. 7352. Problem 278 7353. =========== 7354. 7355. 7356. Given the values of integers 1 < a < a <... < a[n], consider the 7357. linear combination 7358. qa + qa + ... + q[n]a[n] = b, using only integer values q[k] ≥ 7359. 0. 7360. 7361. Note that for a given set of a[k], it may be that not all values of b are 7362. possible. 7363. For instance, if a = 5 and a = 7, there are no q ≥ 0 and q ≥ 0 7364. such that b could be 7365. 1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23. 7366. In fact, 23 is the largest impossible value of b for a = 5 and a = 7367. 7. 7368. We therefore call f(5, 7) = 23. 7369. Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195. 7370. 7371. Find ∑ f(p*q,p*r,q*r), where p, q and r are prime numbers and p < q < r < 7372. 5000. 7373. 7374. 7375. Answer: 7e680606b5e9890a19894dbdbbbd102a 7376. 7377. 7378. Problem 279 7379. =========== 7380. 7381. 7382. How many triangles are there with integral sides, at least one integral 7383. angle (measured in degrees), and a perimeter that does not exceed 10^8? 7384. 7385. 7386. Answer: 1f51455a8180fdeeb21285dfb6cba45f 7387. 7388. 7389. Problem 280 7390. =========== 7391. 7392. 7393. A laborious ant walks randomly on a 5x5 grid. The walk starts from the 7394. central square. At each step, the ant moves to an adjacent square at 7395. random, without leaving the grid; thus there are 2, 3 or 4 possible moves 7396. at each step depending on the ant's position. 7397. 7398. At the start of the walk, a seed is placed on each square of the lower 7399. row. When the ant isn't carrying a seed and reaches a square of the lower 7400. row containing a seed, it will start to carry the seed. The ant will drop 7401. the seed on the first empty square of the upper row it eventually reaches. 7402. 7403. What's the expected number of steps until all seeds have been dropped in 7404. the top row? 7405. Give your answer rounded to 6 decimal places. 7406. 7407. 7408. Answer: 27f07f04d1908e5ce4fa6eac09881cc2 7409. 7410. 7411. Problem 281 7412. =========== 7413. 7414. 7415. You are given a pizza (perfect circle) that has been cut into m·n equal 7416. pieces and you want to have exactly one topping on each slice. 7417. 7418. Let f(m,n) denote the number of ways you can have toppings on the pizza 7419. with m different toppings (m ≥ 2), using each topping on exactly n slices 7420. (n ≥ 1). 7421. Reflections are considered distinct, rotations are not. 7422. 7423. Thus, for instance, f(2,1) = 1, f(2,2) = f(3,1) = 2 and f(3,2) = 16. 7424. f(3,2) is shown below: 7425. 7426. Find the sum of all f(m,n) such that f(m,n) ≤ 10^15. 7427. 7428. 7429. p_281_pizza.gif 7430. Answer: ceee6ced9d64aad844310c8ce2aae2b7 7431. 7432. 7433. Problem 282 7434. =========== 7435. 7436. 7437. For non-negative integers m, n, the Ackermann function A(m, n) is defined 7438. as follows: 7439. 7440. For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125. 7441. 7442. Find A(n, n) and give your answer mod 14^8. 7443. 7444. 7445. p_282_formula.gif 7446. Answer: a1cc665e127af4e907e13087ee777bd5 7447. 7448. 7449. Problem 283 7450. =========== 7451. 7452. 7453. Consider the triangle with sides 6, 8 and 10. It can be seen that the 7454. perimeter and the area are both equal to 24. So the area/perimeter ratio 7455. is equal to 1. 7456. Consider also the triangle with sides 13, 14 and 15. The perimeter equals 7457. 42 while the area is equal to 84. So for this triangle the area/perimeter 7458. ratio is equal to 2. 7459. 7460. Find the sum of the perimeters of all integer sided triangles for which 7461. the area/perimeter ratios are equal to positive integers not exceeding 7462. 1000. 7463. 7464. 7465. Answer: 08afda4bc05c8f3ef71c9ffea1ddc0c8 7466. 7467. 7468. Problem 284 7469. =========== 7470. 7471. 7472. The 3-digit number 376 in the decimal numbering system is an example of 7473. numbers with the special property that its square ends with the same 7474. digits: 376^2 = 141376. Let's call a number with this property a steady 7475. square. 7476. 7477. Steady squares can also be observed in other numbering systems. In the 7478. base 14 numbering system, the 3-digit number c37 is also a steady square: 7479. c37^2 = aa0c37, and the sum of its digits is c+3+7=18 in the same 7480. numbering system. The letters a, b, c and d are used for the 10, 11, 12 7481. and 13 digits respectively, in a manner similar to the hexadecimal 7482. numbering system. 7483. 7484. For 1 ≤ n ≤ 9, the sum of the digits of all the n-digit steady squares in 7485. the base 14 numbering system is 2d8 (582 decimal). Steady squares with 7486. leading 0's are not allowed. 7487. 7488. Find the sum of the digits of all the n-digit steady squares in the base 7489. 14 numbering system for 7490. 1 ≤ n ≤ 10000 (decimal) and give your answer in the base 14 system using 7491. lower case letters where necessary. 7492. 7493. 7494. Answer: aff724582e583649876f518f9b340a69 7495. 7496. 7497. Problem 285 7498. =========== 7499. 7500. 7501. Albert chooses a positive integer k, then two real numbers a, b are 7502. randomly chosen in the interval [0,1] with uniform distribution. 7503. The square root of the sum (k·a+1)^2 + (k·b+1)^2 is then computed and 7504. rounded to the nearest integer. If the result is equal to k, he scores k 7505. points; otherwise he scores nothing. 7506. 7507. For example, if k = 6, a = 0.2 and b = 0.85, then 7508. (k·a+1)^2 + (k·b+1)^2 = 42.05. 7509. The square root of 42.05 is 6.484... and when rounded to the nearest 7510. integer, it becomes 6. 7511. This is equal to k, so he scores 6 points. 7512. 7513. It can be shown that if he plays 10 turns with k = 1, k = 2, ..., k = 10, 7514. the expected value of his total score, rounded to five decimal places, is 7515. 10.20914. 7516. 7517. If he plays 10^5 turns with k = 1, k = 2, k = 3, ..., k = 10^5, what is 7518. the expected value of his total score, rounded to five decimal places? 7519. 7520. 7521. Answer: bbae95d0ce2999cae57782c3746aecb6 7522. 7523. 7524. Problem 286 7525. =========== 7526. 7527. 7528. Barbara is a mathematician and a basketball player. She has found that the 7529. probability of scoring a point when shooting from a distance x is exactly 7530. (1 - ^x/[q]), where q is a real constant greater than 50. 7531. 7532. During each practice run, she takes shots from distances x = 1, x = 2, 7533. ..., x = 50 and, according to her records, she has precisely a 2 % chance 7534. to score a total of exactly 20 points. 7535. 7536. Find q and give your answer rounded to 10 decimal places. 7537. 7538. 7539. Answer: cc5a1ef0deabf698733bcef4f1149498 7540. 7541. 7542. Problem 287 7543. =========== 7544. 7545. 7546. The quadtree encoding allows us to describe a 2^N×2^N black and white 7547. image as a sequence of bits (0 and 1). Those sequences are to be read from 7548. left to right like this: 7549. 7550. • the first bit deals with the complete 2^N×2^N region; 7551. • "0" denotes a split: 7552. the current 2^n×2^n region is divided into 4 sub-regions of dimension 7553. 2^n-1×2^n-1, 7554. the next bits contains the description of the top left, top right, 7555. bottom left and bottom right sub-regions - in that order; 7556. • "10" indicates that the current region contains only black pixels; 7557. • "11" indicates that the current region contains only white pixels. 7558. 7559. Consider the following 4×4 image (colored marks denote places where a 7560. split can occur): 7561. 7562. This image can be described by several sequences, for example 7563. :"001010101001011111011010101010", of length 30, or 7564. "0100101111101110", of length 16, which is the minimal sequence for this 7565. image. 7566. 7567. For a positive integer N, define D[N] as the 2^N×2^N image with the 7568. following coloring scheme: 7569. 7570. • the pixel with coordinates x = 0, y = 0 corresponds to the bottom left 7571. pixel, 7572. • if (x - 2^N-1)^2 + (y - 2^N-1)^2 ≤ 2^2N-2 then the pixel is black, 7573. • otherwise the pixel is white. 7574. 7575. What is the length of the minimal sequence describing D ? 7576. 7577. 7578. p_287_quadtree.gif 7579. Answer: 6c2beec8a6c0bc788d5e45c317b0d7ca 7580. 7581. 7582. Problem 288 7583. =========== 7584. 7585. 7586. For any prime p the number N(p,q) is defined byN(p,q) = ∑[n=0 to q] 7587. T[n]*p^n 7588. with T[n] generated by the following random number generator: 7589. 7590. S = 290797 7591. S[n+1] = S[n]^2 mod 50515093 7592. T[n] = S[n] mod p 7593. 7594. Let Nfac(p,q) be the factorial of N(p,q). 7595. Let NF(p,q) be the number of factors p in Nfac(p,q). 7596. 7597. You are given that NF(3,10000) mod 3^20=624955285. 7598. 7599. Find NF(61,10^7) mod 61^10 7600. 7601. 7602. Answer: 192bf4aa33ea85e922d583f60fe99955 7603. 7604. 7605. Problem 289 7606. =========== 7607. 7608. 7609. Let C(x,y) be a circle passing through the points (x, y), (x, y+1), 7610. (x+1, y) and (x+1, y+1). 7611. 7612. For positive integers m and n, let E(m,n) be a configuration which 7613. consists of the m·n circles: 7614. { C(x,y): 0 ≤ x < m, 0 ≤ y < n, x and y are integers } 7615. 7616. An Eulerian cycle on E(m,n) is a closed path that passes through each arc 7617. exactly once. 7618. Many such paths are possible on E(m,n), but we are only interested in 7619. those which are not self-crossing: A non-crossing path just touches itself 7620. at lattice points, but it never crosses itself. 7621. 7622. The image below shows E(3,3) and an example of an Eulerian non-crossing 7623. path. 7624. 7625. Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n). 7626. For example, L(1,2) = 2, L(2,2) = 37 and L(3,3) = 104290. 7627. 7628. Find L(6,10) mod 10^10. 7629. 7630. 7631. p_289_euler.gif 7632. Answer: 9fa32696df356b3d41faa7dd278c88a9 7633. 7634. 7635. Problem 290 7636. =========== 7637. 7638. 7639. How many integers 0 ≤ n < 10^18 have the property that the sum of the 7640. digits of n equals the sum of digits of 137n? 7641. 7642. 7643. Answer: 8246684fec8ece9f0ee3c9898c8c9d6a 7644. 7645. 7646. Problem 291 7647. =========== 7648. 7649. 7650. A prime number p is called a Panaitopol prime if for some positive 7651. integers 7652. x and y. 7653. 7654. Find how many Panaitopol primes are less than 5×10^15. 7655. 7656. 7657. p_291_formula.gif 7658. Answer: 4037526 7659. 7660. 7661. Problem 292 7662. =========== 7663. 7664. 7665. We shall define a pythagorean polygon to be a convex polygon with the 7666. following properties: 7667. 7668. • there are at least three vertices, 7669. • no three vertices are aligned, 7670. • each vertex has integer coordinates, 7671. • each edge has integer length. 7672. 7673. For a given integer n, define P(n) as the number of distinct pythagorean 7674. polygons for which the perimeter is ≤ n. 7675. Pythagorean polygons should be considered distinct as long as none is a 7676. translation of another. 7677. 7678. You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045. 7679. Find P(120). 7680. 7681. 7682. Answer: 27f50f02ef10f170379b144435e0144b 7683. 7684. 7685. Problem 293 7686. =========== 7687. 7688. 7689. An even positive integer N will be called admissible, if it is a power of 7690. 2 or its distinct prime factors are consecutive primes. 7691. The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48. 7692. 7693. If N is admissible, the smallest integer M > 1 such that N+M is prime, 7694. will be called the pseudo-Fortunate number for N. 7695. 7696. For example, N=630 is admissible since it is even and its distinct prime 7697. factors are the consecutive primes 2,3,5 and 7. 7698. The next prime number after 631 is 641; hence, the pseudo-Fortunate number 7699. for 630 is M=11. 7700. It can also be seen that the pseudo-Fortunate number for 16 is 3. 7701. 7702. Find the sum of all distinct pseudo-Fortunate numbers for admissible 7703. numbers N less than 10^9. 7704. 7705. 7706. Answer: 2209 7707. 7708. 7709. Problem 294 7710. =========== 7711. 7712. 7713. For a positive integer k, define d(k) as the sum of the digits of k in its 7714. usual decimal representation.Thus d(42) = 4+2 = 6. 7715. 7716. For a positive integer n, define S(n) as the number of positive integers k 7717. < 10^n with the following properties : 7718. 7719. • k is divisible by 23 and 7720. • d(k) = 23. 7721. 7722. You are given that S(9) = 263626 and S(42) = 6377168878570056. 7723. 7724. Find S(11^12) and give your answer mod 10^9. 7725. 7726. 7727. Answer: aefe049404a284c7d27fab3887c6c4a2 7728. 7729. 7730. Problem 295 7731. =========== 7732. 7733. 7734. We call the convex area enclosed by two circles a lenticular hole if: 7735. 7736. • The centres of both circles are on lattice points. 7737. • The two circles intersect at two distinct lattice points. 7738. • The interior of the convex area enclosed by both circles does not 7739. contain any lattice points. 7740. 7741. Consider the circles: 7742. C: x^2+y^2=25 7743. C: (x+4)^2+(y-4)^2=1 7744. C: (x-12)^2+(y-4)^2=65 7745. 7746. The circles C, C and C are drawn in the picture below. 7747. 7748. C and C form a lenticular hole, as well as C and C. 7749. 7750. We call an ordered pair of positive real numbers (r, r) a lenticular 7751. pair if there exist two circles with radii r and r that form a 7752. lenticular hole.We can verify that (1, 5) and (5, √65) are the lenticular 7753. pairs of the example above. 7754. 7755. Let L(N) be the number of distinct lenticular pairs (r, r) for which 7756. 0 < r ≤ r ≤ N. 7757. We can verify that L(10) = 30 and L(100) = 3442. 7758. 7759. Find L(100 000). 7760. 7761. 7762. Answer: 5beaace6425205fe879116ee07dae961 7763. 7764. 7765. Problem 296 7766. =========== 7767. 7768. 7769. Given is an integer sided triangle ABC with BC ≤ AC ≤ AB. 7770. k is the angular bisector of angle ACB. 7771. m is the tangent at C to the circumscribed circle of ABC. 7772. n is a line parallel to m through B. 7773. The intersection of n and k is called E. 7774. 7775. How many triangles ABC with a perimeter not exceeding 100 000 exist such 7776. that BE has integral length? 7777. 7778. 7779. Answer: 45986a4405b2dd6c163516319e0c4a1b 7780. 7781. 7782. Problem 297 7783. =========== 7784. 7785. 7786. Each new term in the Fibonacci sequence is generated by adding the 7787. previous two terms. 7788. Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 7789. 34, 55, 89. 7790. 7791. Every positive integer can be uniquely written as a sum of nonconsecutive 7792. terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89. 7793. Such a sum is called the Zeckendorf representation of the number. 7794. 7795. For any integer n>0, let z(n) be the number of terms in the Zeckendorf 7796. representation of n. 7797. Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc. 7798. Also, for 0<n<10^6, ∑ z(n) = 7894453. 7799. 7800. Find ∑ z(n) for 0<n<10^17. 7801. 7802. 7803. Answer: d3fd75f5447698748a826562750a1986 7804. 7805. 7806. Problem 298 7807. =========== 7808. 7809. 7810. Larry and Robin play a memory game involving of a sequence of random 7811. numbers between 1 and 10, inclusive, that are called out one at a time. 7812. Each player can remember up to 5 previous numbers. When the called number 7813. is in a player's memory, that player is awarded a point. If it's not, the 7814. player adds the called number to his memory, removing another number if 7815. his memory is full. 7816. 7817. Both players start with empty memories. Both players always add new missed 7818. numbers to their memory but use a different strategy in deciding which 7819. number to remove: 7820. Larry's strategy is to remove the number that hasn't been called in the 7821. longest time. 7822. Robin's strategy is to remove the number that's been in the memory the 7823. longest time. 7824. 7825. Example game: 7826. 7827. Turn Called Larry's Larry's Robin's Robin's 7828. number memory score memory score 7829. 1 1 1 0 1 0 7830. 2 2 1,2 0 1,2 0 7831. 3 4 1,2,4 0 1,2,4 0 7832. 4 6 1,2,4,6 0 1,2,4,6 0 7833. 5 1 1,2,4,6 1 1,2,4,6 1 7834. 6 8 1,2,4,6,8 1 1,2,4,6,8 1 7835. 7 10 1,4,6,8,10 1 2,4,6,8,10 1 7836. 8 2 1,2,6,8,10 1 2,4,6,8,10 2 7837. 9 4 1,2,4,8,10 1 2,4,6,8,10 3 7838. 10 1 1,2,4,8,10 2 1,4,6,8,10 3 7839. 7840. Denoting Larry's score by L and Robin's score by R, what is the expected 7841. value of |L-R| after 50 turns? Give your answer rounded to eight decimal 7842. places using the format x.xxxxxxxx . 7843. 7844. 7845. Answer: d078fd564995aa2a813a29f44ad79611 7846. 7847. 7848. Problem 299 7849. =========== 7850. 7851. 7852. Four points with integer coordinates are selected: 7853. A(a, 0), B(b, 0), C(0, c) and D(0, d), with 0 < a < b and 0 < c < d. 7854. Point P, also with integer coordinates, is chosen on the line AC so that 7855. the three triangles ABP, CDP and BDP are all similar. 7856. 7857. It is easy to prove that the three triangles can be similar, only if a=c. 7858. 7859. So, given that a=c, we are looking for triplets (a,b,d) such that at least 7860. one point P (with integer coordinates) exists on AC, making the three 7861. triangles ABP, CDP and BDP all similar. 7862. 7863. For example, if (a,b,d)=(2,3,4), it can be easily verified that point 7864. P(1,1) satisfies the above condition. Note that the triplets (2,3,4) and 7865. (2,4,3) are considered as distinct, although point P(1,1) is common for 7866. both. 7867. 7868. If b+d < 100, there are 92 distinct triplets (a,b,d) such that point P 7869. exists. 7870. If b+d < 100 000, there are 320471 distinct triplets (a,b,d) such that 7871. point P exists. 7872. 7873. If b+d < 100 000 000, how many distinct triplets (a,b,d) are there such 7874. that point P exists? 7875. 7876. 7877. p_299_ThreeSimTri.gif 7878. Answer: fb8f093361a6db56c8a1d1661ab229cd 7879. 7880. 7881. Problem 300 7882. =========== 7883. 7884. 7885. In a very simplified form, we can consider proteins as strings consisting 7886. of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH. 7887. For this problem, the orientation of a protein is important; e.g. HPP is 7888. considered distinct from PPH. Thus, there are 2^n distinct proteins 7889. consisting of n elements. 7890. 7891. When one encounters these strings in nature, they are always folded in 7892. such a way that the number of H-H contact points is as large as possible, 7893. since this is energetically advantageous. 7894. As a result, the H-elements tend to accumulate in the inner part, with the 7895. P-elements on the outside. 7896. Natural proteins are folded in three dimensions of course, but we will 7897. only consider protein folding in two dimensions. 7898. 7899. The figure below shows two possible ways that our example protein could be 7900. folded (H-H contact points are shown with red dots). 7901. 7902. The folding on the left has only six H-H contact points, thus it would 7903. never occur naturally. 7904. On the other hand, the folding on the right has nine H-H contact points, 7905. which is optimal for this string. 7906. 7907. Assuming that H and P elements are equally likely to occur in any position 7908. along the string, the average number of H-H contact points in an optimal 7909. folding of a random protein string of length 8 turns out to be 7910. 850 / 2^8=3.3203125. 7911. 7912. What is the average number of H-H contact points in an optimal folding of 7913. a random protein string of length 15? 7914. Give your answer using as many decimal places as necessary for an exact 7915. result. 7916. 7917. 7918. p_300_protein.gif 7919. Answer: 5a0d6315bc18279c46a1fb8cbd2f16b5 7920. 7921. 7922. Problem 301 7923. =========== 7924. 7925. 7926. Nim is a game played with heaps of stones, where two players take it in 7927. turn to remove any number of stones from any heap until no stones remain. 7928. 7929. We'll consider the three-heap normal-play version of Nim, which works as 7930. follows: 7931. - At the start of the game there are three heaps of stones. 7932. - On his turn the player removes any positive number of stones from any 7933. single heap. 7934. - The first player unable to move (because no stones remain) loses. 7935. 7936. If (n,n,n) indicates a Nim position consisting of heaps of size 7937. n, n and n then there is a simple function X(n,n,n) — 7938. that you may look up or attempt to deduce for yourself — that returns: 7939. 7940. • zero if, with perfect strategy, the player about to move will 7941. eventually lose; or 7942. • non-zero if, with perfect strategy, the player about to move will 7943. eventually win. 7944. 7945. For example X(1,2,3) = 0 because, no matter what the current player does, 7946. his opponent can respond with a move that leaves two heaps of equal size, 7947. at which point every move by the current player can be mirrored by his 7948. opponent until no stones remain; so the current player loses. To 7949. illustrate: 7950. - current player moves to (1,2,1) 7951. - opponent moves to (1,0,1) 7952. - current player moves to (0,0,1) 7953. - opponent moves to (0,0,0), and so wins. 7954. 7955. For how many positive integers n ≤ 2^30 does X(n,2n,3n) = 0 ? 7956. 7957. 7958. Answer: 2178309 7959. 7960. 7961. Problem 302 7962. =========== 7963. 7964. 7965. A positive integer n is powerful if p^2 is a divisor of n for every prime 7966. factor p in n. 7967. 7968. A positive integer n is a perfect power if n can be expressed as a power 7969. of another positive integer. 7970. 7971. A positive integer n is an Achilles number if n is powerful but not a 7972. perfect power. For example, 864 and 1800 are Achilles numbers: 864 = 7973. 2^5·3^3 and 1800 = 2^3·3^2·5^2. 7974. 7975. We shall call a positive integer S a Strong Achilles number if both S and 7976. φ(S) are Achilles numbers.^1 7977. For example, 864 is a Strong Achilles number: φ(864) = 288 = 2^5·3^2. 7978. However, 1800 isn't a Strong Achilles number because: φ(1800) = 480 = 7979. 2^5·3^1·5^1. 7980. 7981. There are 7 Strong Achilles numbers below 10^4 and 656 below 10^8. 7982. 7983. How many Strong Achilles numbers are there below 10^18? 7984. 7985. ^1 φ denotes Euler's totient function. 7986. 7987. 7988. Answer: 1170060 7989. 7990. 7991. Problem 303 7992. =========== 7993. 7994. 7995. For a positive integer n, define f(n) as the least positive multiple of n 7996. that, written in base 10, uses only digits ≤ 2. 7997. 7998. Thus f(2)=2, f(3)=12, f(7)=21, f(42)=210, f(89)=1121222. 7999. 8000. Also, . 8001. 8002. Find . 8003. 8004. 8005. Answer: b904a0b3d922e628a828e744ee7d3a60 8006. 8007. 8008. Problem 304 8009. =========== 8010. 8011. 8012. For any positive integer n the function next_prime(n) returns the smallest 8013. prime p 8014. such that p>n. 8015. 8016. The sequence a(n) is defined by: 8017. a(1)=next_prime(10^14) and a(n)=next_prime(a(n-1)) for n>1. 8018. 8019. The fibonacci sequence f(n) is defined by:f(0)=0, f(1)=1 and 8020. f(n)=f(n-1)+f(n-2) for n>1. 8021. 8022. The sequence b(n) is defined as f(a(n)). 8023. 8024. Find ∑b(n) for 1≤n≤100 000. Give your answer mod 1234567891011. 8025. 8026. 8027. Answer: 499427a3e4bf9ad34a6df3056604b4c1 8028. 8029. 8030. Problem 305 8031. =========== 8032. 8033. 8034. Let's call S the (infinite) string that is made by concatenating the 8035. consecutive positive integers (starting from 1) written down in base 10. 8036. Thus, S = 1234567891011121314151617181920212223242... 8037. 8038. It's easy to see that any number will show up an infinite number of times 8039. in S. 8040. 8041. Let's call f(n) the starting position of the n^th occurrence of n in S. 8042. For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365. 8043. 8044. Find ∑f(3^k) for 1≤k≤13. 8045. 8046. 8047. Answer: 9def85298f598867d361e4afca8cdd96 8048. 8049. 8050. Problem 306 8051. =========== 8052. 8053. 8054. The following game is a classic example of Combinatorial Game Theory: 8055. 8056. Two players start with a strip of n white squares and they take alternate 8057. turns. 8058. On each turn, a player picks two contiguous white squares and paints them 8059. black. 8060. The first player who cannot make a move loses. 8061. 8062. • If n = 1, there are no valid moves, so the first player loses 8063. automatically. 8064. • If n = 2, there is only one valid move, after which the second player 8065. loses. 8066. • If n = 3, there are two valid moves, but both leave a situation where 8067. the second player loses. 8068. • If n = 4, there are three valid moves for the first player; she can 8069. win the game by painting the two middle squares. 8070. • If n = 5, there are four valid moves for the first player (shown below 8071. in red); but no matter what she does, the second player (blue) wins. 8072. 8073. So, for 1 ≤ n ≤ 5, there are 3 values of n for which the first player can 8074. force a win. 8075. Similarly, for 1 ≤ n ≤ 50, there are 40 values of n for which the first 8076. player can force a win. 8077. 8078. For 1 ≤ n ≤ 1 000 000, how many values of n are there for which the first 8079. player can force a win? 8080. 8081. 8082. p_306_pstrip.gif 8083. Answer: 852938 8084. 8085. 8086. Problem 307 8087. =========== 8088. 8089. 8090. k defects are randomly distributed amongst n integrated-circuit chips 8091. produced by a factory (any number of defects may be found on a chip and 8092. each defect is independent of the other defects). 8093. 8094. Let p(k,n) represent the probability that there is a chip with at least 3 8095. defects. 8096. For instance p(3,7) ≈ 0.0204081633. 8097. 8098. Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal 8099. places in the form 0.abcdefghij 8100. 8101. 8102. Answer: 0c49094fa750365e13bb20ec4a158b6d 8103. 8104. 8105. Problem 308 8106. =========== 8107. 8108. 8109. A program written in the programming language Fractran consists of a list 8110. of fractions. 8111. 8112. The internal state of the Fractran Virtual Machine is a positive integer, 8113. which is initially set to a seed value. Each iteration of a Fractran 8114. program multiplies the state integer by the first fraction in the list 8115. which will leave it an integer. 8116. 8117. For example, one of the Fractran programs that John Horton Conway wrote 8118. for prime-generation consists of the following 14 fractions: 8119. 8120. 17 , 78 , 19 , 23 , 29 , 77 , 95 , 77 , 1 , 11 , 13 , 15 , 1 , 55 . 8121. 91 85 51 38 33 29 23 19 17 13 11 2 7 1 8122. 8123. Starting with the seed integer 2, successive iterations of the program 8124. produce the sequence: 8125. 15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544, 8126. 32, 240, ... 8127. 8128. The powers of 2 that appear in this sequence are 2^2, 2^3, 2^5, ... 8129. It can be shown that all the powers of 2 in this sequence have prime 8130. exponents and that all the primes appear as exponents of powers of 2, in 8131. proper order! 8132. 8133. If someone uses the above Fractran program to solve Project Euler Problem 8134. 7 (find the 10001^st prime), how many iterations would be needed until the 8135. program produces 2^10001st prime ? 8136. 8137. 8138. Answer: 43e736dfc6478a52653814248a71771d 8139. 8140. 8141. Problem 309 8142. =========== 8143. 8144. 8145. In the classic "Crossing Ladders" problem, we are given the lengths x and 8146. y of two ladders resting on the opposite walls of a narrow, level street. 8147. We are also given the height h above the street where the two ladders 8148. cross and we are asked to find the width of the street (w). 8149. 8150. Here, we are only concerned with instances where all four variables are 8151. positive integers. 8152. For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56. 8153. 8154. In fact, for integer values x, y, h and 0 < x < y < 200, there are only 8155. five triplets (x,y,h) producing integer solutions for w: 8156. (70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175, 8157. 40). 8158. 8159. For integer values x, y, h and 0 < x < y < 1 000 000, how many triplets 8160. (x,y,h) produce integer solutions for w? 8161. 8162. 8163. p_309_ladders.gif 8164. Answer: 210139 8165. 8166. 8167. Problem 310 8168. =========== 8169. 8170. 8171. Alice and Bob play the game Nim Square. 8172. Nim Square is just like ordinary three-heap normal play Nim, but the 8173. players may only remove a square number of stones from a heap. 8174. The number of stones in the three heaps is represented by the ordered 8175. triple (a,b,c). 8176. If 0≤a≤b≤c≤29 then the number of losing positions for the next player is 8177. 1160. 8178. 8179. Find the number of losing positions for the next player if 0≤a≤b≤c≤100 8180. 000. 8181. 8182. 8183. Answer: 6b94f848996393eef163add4d17360c7 8184. 8185. 8186. Problem 311 8187. =========== 8188. 8189. 8190. ABCD is a convex, integer sided quadrilateral with 1 ≤ AB < BC < CD < AD. 8191. BD has integer length. O is the midpoint of BD. AO has integer length. 8192. We'll call ABCD a biclinic integral quadrilateral if AO = CO ≤ BO = DO. 8193. 8194. For example, the following quadrilateral is a biclinic integral 8195. quadrilateral: 8196. AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23. 8197. 8198. Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD 8199. that satisfy AB^2+BC^2+CD^2+AD^2 ≤ N. 8200. We can verify that B(10 000) = 49 and B(1 000 000) = 38239. 8201. 8202. Find B(10 000 000 000). 8203. 8204. 8205. p_311_biclinic.gif 8206. Answer: 36115d4f7dc07eea106d78e8431868e6 8207. 8208. 8209. Problem 312 8210. =========== 8211. 8212. 8213. - A Sierpiński graph of order-1 (S) is an equilateral triangle. 8214. - S[n+1] is obtained from S[n] by positioning three copies of S[n] so that 8215. every pair of copies has one common corner. 8216. 8217. Let C(n) be the number of cycles that pass exactly once through all the 8218. vertices of S[n]. 8219. For example, C(3) = 8 because eight such cycles can be drawn on S, as 8220. shown below: 8221. 8222. It can also be verified that : 8223. C(1) = C(2) = 1 8224. C(5) = 71328803586048 8225. C(10 000) mod 10^8 = 37652224 8226. C(10 000) mod 13^8 = 617720485 8227. 8228. Find C(C(C(10 000))) mod 13^8. 8229. 8230. 8231. p_312_sierpinskyAt.gif 8232. p_312_sierpinsky8t.gif 8233. Answer: 535113d1a81f421fe814d48205dac570 8234. 8235. 8236. Problem 313 8237. =========== 8238. 8239. 8240. In a sliding game a counter may slide horizontally or vertically into an 8241. empty space. The objective of the game is to move the red counter from the 8242. top left corner of a grid to the bottom right corner; the space always 8243. starts in the bottom right corner. For example, the following sequence of 8244. pictures show how the game can be completed in five moves on a 2 by 2 8245. grid. 8246. 8247. Let S(m,n) represent the minimum number of moves to complete the game on 8248. an m by n grid. For example, it can be verified that S(5,4) = 25. 8249. 8250. There are exactly 5482 grids for which S(m,n) = p^2, where p < 100 is 8251. prime. 8252. 8253. How many grids does S(m,n) = p^2, where p < 10^6 is prime? 8254. 8255. 8256. p_313_sliding_game_1.gif 8257. p_313_sliding_game_2.gif 8258. Answer: 2468d42fa1c7f61547ce71c9826218ea 8259. 8260. 8261. Problem 314 8262. =========== 8263. 8264. 8265. The moon has been opened up, and land can be obtained for free, but there 8266. is a catch. You have to build a wall around the land that you stake out, 8267. and building a wall on the moon is expensive. Every country has been 8268. allotted a 500 m by 500 m square area, but they will possess only that 8269. area which they wall in. 251001 posts have been placed in a rectangular 8270. grid with 1 meter spacing. The wall must be a closed series of straight 8271. lines, each line running from post to post. 8272. 8273. The bigger countries of course have built a 2000 m wall enclosing the 8274. entire 250 000 m^2 area. The Duchy of Grand Fenwick, has a tighter 8275. budget, and has asked you (their Royal Programmer) to compute what shape 8276. would get best maximum enclosed-area/wall-length ratio. 8277. 8278. You have done some preliminary calculations on a sheet of paper.For a 2000 8279. meter wall enclosing the 250 000 m^2 area theenclosed-area/wall-length 8280. ratio is 125. 8281. Although not allowed , but to get an idea if this is anything better: if 8282. you place a circle inside the square area touching the four sides the area 8283. will be equal to π*250^2 m^2 and the perimeter will be π*500 m, so the 8284. enclosed-area/wall-length ratio will also be 125. 8285. 8286. However, if you cut off from the square four triangles with sides 75 m, 75 8287. m and 75√2 m the total area becomes 238750 m^2 and the perimeter becomes 8288. 1400+300√2 m. So this gives an enclosed-area/wall-length ratio of 130.87, 8289. which is significantly better. 8290. 8291. Find the maximum enclosed-area/wall-length ratio. 8292. Give your answer rounded to 8 places behind the decimal point in the form 8293. abc.defghijk. 8294. 8295. 8296. Visible links 8297. 1. http://en.wikipedia.org/wiki/Grand_Fenwick 8298. p_314_landgrab.gif 8299. Answer: aa457cae6f67945d50683a85a9b70230 8300. 8301. 8302. Problem 315 8303. =========== 8304. 8305. 8306. Sam and Max are asked to transform two digital clocks into two "digital 8307. root" clocks. 8308. A digital root clock is a digital clock that calculates digital roots step 8309. by step. 8310. 8311. When a clock is fed a number, it will show it and then it will start the 8312. calculation, showing all the intermediate values until it gets to the 8313. result. 8314. For example, if the clock is fed the number 137, it will show: "137" → 8315. "11" → "2" and then it will go black, waiting for the next number. 8316. 8317. Every digital number consists of some light segments: three horizontal 8318. (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, 8319. bottom-right). 8320. Number "1" is made of vertical top-right and bottom-right, number "4" is 8321. made by middle horizontal and vertical top-left, top-right and 8322. bottom-right. Number "8" lights them all. 8323. 8324. The clocks consume energy only when segments are turned on/off. 8325. To turn on a "2" will cost 5 transitions, while a "7" will cost only 4 8326. transitions. 8327. 8328. Sam and Max built two different clocks. 8329. 8330. Sam's clock is fed e.g. number 137: the clock shows "137", then the panel 8331. is turned off, then the next number ("11") is turned on, then the panel is 8332. turned off again and finally the last number ("2") is turned on and, after 8333. some time, off. 8334. For the example, with number 137, Sam's clock requires: 8335. 8336. "137" : (2 + 5 + 4) × 2 = 22 transitions ("137" on/off). 8337. "11" : (2 + 2) × 2 = 8 transitions ("11" on/off). 8338. "2" : (5) × 2 = 10 transitions ("2" on/off). 8339. 8340. For a grand total of 40 transitions. 8341. 8342. Max's clock works differently. Instead of turning off the whole panel, it 8343. is smart enough to turn off only those segments that won't be needed for 8344. the next number. 8345. For number 137, Max's clock requires: 8346. 8347. 2 + 5 + 4 = 11 transitions ("137" on) 8348. "137" : 7 transitions (to turn off the segments that are not needed for 8349. number "11"). 8350. 0 transitions (number "11" is already turned on correctly) 8351. "11" : 3 transitions (to turn off the first "1" and the bottom part of 8352. the second "1"; 8353. the top part is common with number "2"). 8354. 4 tansitions (to turn on the remaining segments in order to get a 8355. "2" : "2") 8356. 5 transitions (to turn off number "2"). 8357. 8358. For a grand total of 30 transitions. 8359. 8360. Of course, Max's clock consumes less power than Sam's one. 8361. The two clocks are fed all the prime numbers between A = 10^7 and B = 8362. 2×10^7. 8363. Find the difference between the total number of transitions needed by 8364. Sam's clock and that needed by Max's one. 8365. 8366. 8367. p_315_clocks.gif 8368. Answer: 13625242 8369. 8370. 8371. Problem 316 8372. =========== 8373. 8374. 8375. Let p = p p p ... be an infinite sequence of random digits, 8376. selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. 8377. It can be seen that p corresponds to the real number 0.p p p .... 8378. It can also be seen that choosing a random real number from the interval 8379. [0,1) is equivalent to choosing an infinite sequence of random digits 8380. selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. 8381. 8382. For any positive integer n with d decimal digits, let k be the smallest 8383. index such that 8384. p[k, ]p[k+1], ...p[k+d-1] are the decimal digits of n, in the same order. 8385. Also, let g(n) be the expected value of k; it can be proven that g(n) is 8386. always finite and, interestingly, always an integer number. 8387. 8388. For example, if n = 535, then 8389. for p = 31415926535897...., we get k = 9 8390. for p = 355287143650049560000490848764084685354..., we get k = 36 8391. etc and we find that g(535) = 1008. 8392. 8393. Given that , find 8394. 8395. Note: represents the floor function. 8396. 8397. p_316_decexp1.gif 8398. p_316_decexp2.gif 8399. p_316_decexp3.gif 8400. Answer: 2495e8f6e9d4cdadbf0411144e7180b9 8401. 8402. 8403. Problem 317 8404. =========== 8405. 8406. 8407. A firecracker explodes at a height of 100 m above level ground. It breaks 8408. into a large number of very small fragments, which move in every 8409. direction; all of them have the same initial velocity of 20 m/s. 8410. 8411. We assume that the fragments move without air resistance, in a uniform 8412. gravitational field with g=9.81 m/s^2. 8413. 8414. Find the volume (in m^3) of the region through which the fragments move 8415. before reaching the ground. Give your answer rounded to four decimal 8416. places. 8417. 8418. 8419. Answer: b0e2bec93bfe598ade5d3d1141f76bdd 8420. 8421. 8422. Problem 318 8423. =========== 8424. 8425. 8426. Consider the real number √2+√3. 8427. When we calculate the even powers of √2+√3we get: 8428. (√2+√3)^2 = 9.898979485566356... 8429. (√2+√3)^4 = 97.98979485566356... 8430. (√2+√3)^6 = 969.998969071069263... 8431. (√2+√3)^8 = 9601.99989585502907... 8432. (√2+√3)^10 = 95049.999989479221... 8433. (√2+√3)^12 = 940897.9999989371855... 8434. (√2+√3)^14 = 9313929.99999989263... 8435. (√2+√3)^16 = 92198401.99999998915... 8436. 8437. It looks like that the number of consecutive nines at the beginning of the 8438. fractional part of these powers is non-decreasing. 8439. In fact it can be proven that the fractional part of (√2+√3)^2n approaches 8440. 1 for large n. 8441. 8442. Consider all real numbers of the form √p+√q with p and q positive integers 8443. and p<q, such that the fractional part of (√p+√q)^2n approaches 1 for 8444. large n. 8445. 8446. Let C(p,q,n) be the number of consecutive nines at the beginning of the 8447. fractional part of 8448. (√p+√q)^2n. 8449. 8450. Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011. 8451. 8452. Find ∑N(p,q) for p+q ≤ 2011. 8453. 8454. 8455. Answer: de358f1c4d6e30c1a4f82c8bc5cedf2d 8456. 8457. 8458. Problem 319 8459. =========== 8460. 8461. 8462. Let x, x,..., x[n] be a sequence of length n such that: 8463. 8464. • x = 2 8465. • for all 1 < i ≤ n : x[i-1] < x[i] 8466. • for all i and j with 1 ≤ i, j ≤ n : (x[i]) ^ j < (x[j] + 1)^i 8467. 8468. There are only five such sequences of length 2, namely:{2,4}, {2,5}, 8469. {2,6}, {2,7} and {2,8}. 8470. There are 293 such sequences of length 5; three examples are given below: 8471. {2,5,11,25,55}, {2,6,14,36,88}, {2,8,22,64,181}. 8472. 8473. Let t(n) denote the number of such sequences of length n. 8474. You are given that t(10) = 86195 and t(20) = 5227991891. 8475. 8476. Find t(10^10) and give your answer modulo 10^9. 8477. 8478. 8479. Answer: d346ab7d128ee0402820edf5fe4aed30 8480. 8481. 8482. Problem 320 8483. =========== 8484. 8485. 8486. Let N(i) be the smallest integer n such that n! is divisible by 8487. (i!)^1234567890 8488. 8489. Let S(u)=∑N(i) for 10 ≤ i ≤ u. 8490. 8491. S(1000)=614538266565663. 8492. 8493. Find S(1 000 000) mod 10^18. 8494. 8495. 8496. Answer: 8426f939c3ee410a8c4d43886ef77ccb 8497. 8498. 8499. Problem 321 8500. =========== 8501. 8502. 8503. A horizontal row comprising of 2n + 1 squares has n red counters placed at 8504. one end and n blue counters at the other end, being separated by a single 8505. empty square in the centre. For example, when n = 3. 8506. 8507. A counter can move from one square to the next (slide) or can jump over 8508. another counter (hop) as long as the square next to that counter is 8509. unoccupied. 8510. 8511. Let M(n) represent the minimum number of moves/actions to completely 8512. reverse the positions of the coloured counters; that is, move all the red 8513. counters to the right and all the blue counters to the left. 8514. 8515. It can be verified M(3) = 15, which also happens to be a triangle number. 8516. 8517. If we create a sequence based on the values of n for which M(n) is a 8518. triangle number then the first five terms would be: 8519. 1, 3, 10, 22, and 63, and their sum would be 99. 8520. 8521. Find the sum of the first forty terms of this sequence. 8522. 8523. 8524. p_321_swapping_counters_1.gif 8525. p_321_swapping_counters_2.gif 8526. Answer: 6d87412130312b01a999225a5fe689b1 8527. 8528. 8529. Problem 322 8530. =========== 8531. 8532. 8533. Let T(m, n) be the number of the binomial coefficients ^iC[n] that are 8534. divisible by 10 for n ≤ i < m(i, m and n are positive integers). 8535. You are given that T(10^9, 10^7-10) = 989697000. 8536. 8537. Find T(10^18, 10^12-10). 8538. 8539. 8540. Answer: a75af9d717fa592487fb45e7552204a8 8541. 8542. 8543. Problem 323 8544. =========== 8545. 8546. 8547. Let y, y, y,... be a sequence of random unsigned 32 bit integers 8548. (i.e. 0 ≤ y[i] < 2^32, every value equally likely). 8549. 8550. For the sequence x[i] the following recursion is given: 8551. 8552. • x = 0 and 8553. • x[i] = x[i-1] | y[i-1], for i > 0. ( | is the bitwise-OR operator) 8554. 8555. It can be seen that eventually there will be an index N such that x[i] = 8556. 2^32 -1 (a bit-pattern of all ones) for all i ≥ N. 8557. 8558. Find the expected value of N. 8559. Give your answer rounded to 10 digits after the decimal point. 8560. 8561. 8562. Answer: c8f8a7ab17a87f1b17a1f4a86c984ea7 8563. 8564. 8565. Problem 324 8566. =========== 8567. 8568. 8569. Let f(n) represent the number of ways one can fill a 3×3×n tower with 8570. blocks of 2×1×1. 8571. You're allowed to rotate the blocks in any way you like; however, 8572. rotations, reflections etc of the tower itself are counted as distinct. 8573. 8574. For example (with q = 100000007) : 8575. f(2) = 229, 8576. f(4) = 117805, 8577. f(10) mod q = 96149360, 8578. f(10^3) mod q = 24806056, 8579. f(10^6) mod q = 30808124. 8580. 8581. Find f(10^10000) mod 100000007. 8582. 8583. 8584. Answer: 96972774 8585. 8586. 8587. Problem 325 8588. =========== 8589. 8590. 8591. A game is played with two piles of stones and two players. At her turn, a 8592. player removes a number of stones from the larger pile. The number of 8593. stones she removes must be a positive multiple of the number of stones in 8594. the smaller pile. 8595. 8596. E.g., let the ordered pair(6,14) describe a configuration with 6 stones in 8597. the smaller pile and 14 stones in the larger pile, then the first player 8598. can remove 6 or 12 stones from the larger pile. 8599. 8600. The player taking all the stones from a pile wins the game. 8601. 8602. A winning configuration is one where the first player can force a win. For 8603. example, (1,5), (2,6) and (3,12) are winning configurations because the 8604. first player can immediately remove all stones in the second pile. 8605. 8606. A losing configuration is one where the second player can force a win, no 8607. matter what the first player does. For example, (2,3) and (3,4) are losing 8608. configurations: any legal move leaves a winning configuration for the 8609. second player. 8610. 8611. Define S(N) as the sum of (x[i]+y[i]) for all losing configurations 8612. (x[i],y[i]), 0 < x[i] < y[i] ≤ N. We can verify that S(10) = 211 and 8613. S(10^4) = 230312207313. 8614. 8615. Find S(10^16) mod 7^10. 8616. 8617. 8618. Answer: 54672965 8619. 8620. 8621. Problem 326 8622. =========== 8623. 8624. 8625. Let a[n] be a sequence recursively defined by: . 8626. 8627. So the first 10 elements of a[n] are: 1,1,0,3,0,3,5,4,1,9. 8628. 8629. Let f(N,M) represent the number of pairs (p,q) such that: 8630. 8631. It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and 8632. (9,10). 8633. 8634. You are also given that f(10^4,10^3)=97158. 8635. 8636. Find f(10^12,10^6). 8637. 8638. 8639. p_326_formula1.gif 8640. p_326_formula2.gif 8641. Answer: d95dff1a5ceee0064993d98defdd603e 8642. 8643. 8644. Problem 327 8645. =========== 8646. 8647. 8648. A series of three rooms are connected to each other by automatic doors. 8649. 8650. Each door is operated by a security card. Once you enter a room the door 8651. automatically closes and that security card cannot be used again. A 8652. machine at the start will dispense an unlimited number of cards, but each 8653. room (including the starting room) contains scanners and if they detect 8654. that you are holding more than three security cards or if they detect an 8655. unattended security card on the floor, then all the doors will become 8656. permanently locked. However, each room contains a box where you may safely 8657. store any number of security cards for use at a later stage. 8658. 8659. If you simply tried to travel through the rooms one at a time then as you 8660. entered room 3 you would have used all three cards and would be trapped in 8661. that room forever! 8662. 8663. However, if you make use of the storage boxes, then escape is possible. 8664. For example, you could enter room 1 using your first card, place one card 8665. in the storage box, and use your third card to exit the room back to the 8666. start. Then after collecting three more cards from the dispensing machine 8667. you could use one to enter room 1 and collect the card you placed in the 8668. box a moment ago. You now have three cards again and will be able to 8669. travel through the remaining three doors. This method allows you to travel 8670. through all three rooms using six security cards in total. 8671. 8672. It is possible to travel through six rooms using a total of 123 security 8673. cards while carrying a maximum of 3 cards. 8674. 8675. Let C be the maximum number of cards which can be carried at any time. 8676. 8677. Let R be the number of rooms to travel through. 8678. 8679. Let M(C,R) be the minimum number of cards required from the dispensing 8680. machine to travel through R rooms carrying up to a maximum of C cards at 8681. any time. 8682. 8683. For example, M(3,6)=123 and M(4,6)=23. 8684. And, ΣM(C,6)=146 for 3 ≤ C ≤ 4. 8685. 8686. You are given that ΣM(C,10)=10382 for 3 ≤ C ≤ 10. 8687. 8688. Find ΣM(C,30) for 3 ≤ C ≤ 40. 8689. 8690. 8691. p_327_rooms_of_doom.gif 8692. Answer: 2cd4c0ad8a00c5be99802188ee2628fb 8693. 8694. 8695. Problem 328 8696. =========== 8697. 8698. 8699. We are trying to find a hidden number selected from the set of integers 8700. {1, 2, ..., n} by asking questions. Each number (question) we ask, has a 8701. cost equal to the number asked and we get one of three possible answers: 8702. 8703. • "Your guess is lower than the hidden number", or 8704. • "Yes, that's it!", or 8705. • "Your guess is higher than the hidden number". 8706. 8707. Given the value of n, an optimal strategy minimizes the total cost (i.e. 8708. the sum of all the questions asked) for the worst possible case. E.g. 8709. 8710. If n=3, the best we can do is obviously to ask the number "2". The answer 8711. will immediately lead us to find the hidden number (at a total cost = 2). 8712. 8713. If n=8, we might decide to use a "binary search" type of strategy: Our 8714. first question would be "4" and if the hidden number is higher than 4 we 8715. will need one or two additional questions. 8716. Let our second question be "6". If the hidden number is still higher than 8717. 6, we will need a third question in order to discriminate between 7 and 8. 8718. Thus, our third question will be "7" and the total cost for this 8719. worst-case scenario will be 4+6+7=17. 8720. 8721. We can improve considerably the worst-case cost for n=8, by asking "5" as 8722. our first question. 8723. If we are told that the hidden number is higher than 5, our second 8724. question will be "7", then we'll know for certain what the hidden number 8725. is (for a total cost of 5+7=12). 8726. If we are told that the hidden number is lower than 5, our second question 8727. will be "3" and if the hidden number is lower than 3 our third question 8728. will be "1", giving a total cost of 5+3+1=9. 8729. Since 12>9, the worst-case cost for this strategy is 12. That's better 8730. than what we achieved previously with the "binary search" strategy; it is 8731. also better than or equal to any other strategy. 8732. So, in fact, we have just described an optimal strategy for n=8. 8733. 8734. Let C(n) be the worst-case cost achieved by an optimal strategy for n, as 8735. described above. 8736. Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12. 8737. Similarly, C(100) = 400 and C(n) = 17575. 8738. 8739. Find C(n). 8740. 8741. 8742. p_328_sum1.gif 8743. p_328_sum2.gif 8744. Answer: 92a3220ad5b17a562c039e6e93d6df90 8745. 8746. 8747. Problem 329 8748. =========== 8749. 8750. 8751. Susan has a prime frog. 8752. Her frog is jumping around over 500 squares numbered 1 to 500.He can only 8753. jump one square to the left or to the right, with equal probability, and 8754. he cannot jump outside the range [1;500]. 8755. (if it lands at either end, it automatically jumps to the only available 8756. square on the next move.) 8757. 8758. When he is on a square with a prime number on it, he croaks 'P' (PRIME) 8759. with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before 8760. jumping to the next square. 8761. When he is on a square with a number on it that is not a prime he croaks 8762. 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping 8763. to the next square. 8764. 8765. Given that the frog's starting position is random with the same 8766. probability for every square, and given that she listens to his first 15 8767. croaks, what is the probability that she hears the sequence 8768. PPPPNNPPPNPPNPN? 8769. 8770. Give your answer as a fraction p/q in reduced form. 8771. 8772. Answer: e392a8b1b053c83e68663e08456bb392 8773. 8774. 8775. Problem 330 8776. =========== 8777. 8778. An infinite sequence of real numbers a(n) is defined for all integers n as 8779. follows: 8780. 8781. For example, 8782. 8783. a(0) = 1 + 1 + 1 + ... = e − 1 8784. 1! 2! 3! 8785. 8786. a(1) = e − 1 + 1 + 1 + ... = 2e − 3 8787. 1! 2! 3! 8788. 8789. a(2) = 2e − 3 + e − 1 + 1 + ... = 7 e − 6 8790. 1! 2! 3! 2 8791. 8792. with e = 2.7182818... being Euler's constant. 8793. 8794. It can be shown that a(n) is of A(n) e + B(n) for integers A(n) and B(n). 8795. the form n! 8796. 8797. For example a(10) = 328161643 e − 652694486 . 8798. 10! 8799. 8800. Find A(10^9) + B(10^9) and give your answer mod 77 777 777. 8801. 8802. 8803. p_330_formula.gif 8804. Answer: 15955822 8805. 8806. 8807. Problem 331 8808. =========== 8809. 8810. 8811. N×N disks are placed on a square game board. Each disk has a black side 8812. and white side. 8813. 8814. At each turn, you may choose a disk and flip all the disks in the same row 8815. and the same column as this disk: thus 2×N-1 disks are flipped. The game 8816. ends when all disks show their white side. The following example shows a 8817. game on a 5×5 board. 8818. 8819. It can be proven that 3 is the minimal number of turns to finish this 8820. game. 8821. 8822. The bottom left disk on the N×N board has coordinates (0,0); 8823. the bottom right disk has coordinates (N-1,0) and the top left disk has 8824. coordinates (0,N-1). 8825. 8826. Let C[N] be the following configuration of a board with N×N disks: 8827. A disk at (x,y) satisfying , shows its black side; otherwise, it shows its 8828. white side. C is shown above. 8829. 8830. Let T(N) be the minimal number of turns to finish a game starting from 8831. configuration C[N] or 0 if configuration C[N] is unsolvable. 8832. We have shown that T(5)=3. You are also given that T(10)=29 and T(1 8833. 000)=395253. 8834. 8835. Find . 8836. 8837. 8838. p_331_crossflips3.gif 8839. p_331_crossflips1.gif 8840. p_331_crossflips2.gif 8841. Answer: b609ccc578e71db9de0524fff94e1b70 8842. 8843. 8844. Problem 332 8845. =========== 8846. 8847. 8848. A spherical triangle is a figure formed on the surface of a sphere by 8849. three great circular arcs intersecting pairwise in three vertices. 8850. 8851. Let C(r) be the sphere with the centre (0,0,0) and radius r. 8852. Let Z(r) be the set of points on the surface of C(r) with integer 8853. coordinates. 8854. Let T(r) be the set of spherical triangles with vertices in 8855. Z(r).Degenerate spherical triangles, formed by three points on the same 8856. great arc, are not included in T(r). 8857. Let A(r) be the area of the smallest spherical triangle in T(r). 8858. 8859. For example A(14) is 3.294040 rounded to six decimal places. 8860. 8861. Find A(r). Give your answer rounded to six decimal places. 8862. 8863. 8864. p_332_spherical.jpg 8865. p_332_sum.gif 8866. Answer: c2ae53ebfb15db373cfe5d71078ea1ca 8867. 8868. 8869. Problem 333 8870. =========== 8871. 8872. 8873. All positive integers can be partitioned in such a way that each and every 8874. term of the partition can be expressed as 2^ix3^j, where i,j ≥ 0. 8875. 8876. Let's consider only those such partitions where none of the terms can 8877. divide any of the other terms. 8878. For example, the partition of 17 = 2 + 6 + 9 = (2^1x3^0 + 2^1x3^1 + 8879. 2^0x3^2) would not be valid since 2 can divide 6. Neither would the 8880. partition 17 = 16 + 1 = (2^4x3^0 + 2^0x3^0) since 1 can divide 16. The 8881. only valid partition of 17 would be 8 + 9 = (2^3x3^0 + 2^0x3^2). 8882. 8883. Many integers have more than one valid partition, the first being 11 8884. having the following two partitions. 8885. 11 = 2 + 9 = (2^1x3^0 + 2^0x3^2) 8886. 11 = 8 + 3 = (2^3x3^0 + 2^0x3^1) 8887. 8888. Let's define P(n) as the number of valid partitions of n. For example, 8889. P(11) = 2. 8890. 8891. Let's consider only the prime integers q which would have a single valid 8892. partition such as P(17). 8893. 8894. The sum of the primes q <100 such that P(q)=1 equals 233. 8895. 8896. Find the sum of the primes q <1000000 such that P(q)=1. 8897. 8898. 8899. Answer: 3053105 8900. 8901. 8902. Problem 334 8903. =========== 8904. 8905. 8906. In Plato's heaven, there exist an infinite number of bowls in a straight 8907. line. 8908. Each bowl either contains some or none of a finite number of beans. 8909. A child plays a game, which allows only one kind of move: removing two 8910. beans from any bowl, and putting one in each of the two adjacent bowls. 8911. The game ends when each bowl contains either one or no beans. 8912. 8913. For example, consider two adjacent bowls containing 2 and 3 beans 8914. respectively, all other bowls being empty. The following eight moves will 8915. finish the game: 8916. 8917. You are given the following sequences: 8918. 8919. t = 123456. 8920. 8921. t[i-1] , if t[i-1] is even 8922. t[i] = 2 8923. t[i-1] 926252, if t[i-1] is odd 8924. 2 8925. where ⌊x⌋ is the floor function 8926. and is the bitwise XOR operator. 8927. 8928. b[i] = ( t[i] mod 2^11) + 1. 8929. 8930. The first two terms of the last sequence are b = 289 and b = 145. 8931. If we start with b and b beans in two adjacent bowls, 3419100 moves 8932. would be required to finish the game. 8933. 8934. Consider now 1500 adjacent bowls containing b, b,..., b beans 8935. respectively, all other bowls being empty. Find how many moves it takes 8936. before the game ends. 8937. 8938. 8939. p_334_beans.gif 8940. p_334_cases.gif 8941. p_334_lfloor.gif 8942. p_334_rfloor.gif 8943. p_334_oplus.gif 8944. Answer: 71851da3058acf6b74e90251bdf4aa8f 8945. 8946. 8947. Problem 335 8948. =========== 8949. 8950. 8951. Whenever Peter feels bored, he places some bowls, containing one bean 8952. each, in a circle. After this, he takes all the beans out of a certain 8953. bowl and drops them one by one in the bowls going clockwise. He repeats 8954. this, starting from the bowl he dropped the last bean in, until the 8955. initial situation appears again. For example with 5 bowls he acts as 8956. follows: 8957. 8958. So with 5 bowls it takes Peter 15 moves to return to the initial 8959. situation. 8960. 8961. Let M(x) represent the number of moves required to return to the initial 8962. situation, starting with x bowls. Thus, M(5) = 15. It can also be verified 8963. that M(100) = 10920. 8964. 8965. Find M(2^k+1). Give your answer modulo 7^9. 8966. 8967. 8968. p_335_mancala.gif 8969. p_335_sum.gif 8970. Answer: 5032316 8971. 8972. 8973. Problem 336 8974. =========== 8975. 8976. 8977. A train is used to transport four carriages in the order: ABCD. However, 8978. sometimes when the train arrives to collect the carriages they are not in 8979. the correct order. 8980. To rearrange the carriages they are all shunted on to a large rotating 8981. turntable. After the carriages are uncoupled at a specific point the train 8982. moves off the turntable pulling the carriages still attached with it. The 8983. remaining carriages are rotated 180 degrees. All of the carriages are then 8984. rejoined and this process is repeated as often as necessary in order to 8985. obtain the least number of uses of the turntable. 8986. Some arrangements, such as ADCB, can be solved easily: the carriages are 8987. separated between A and D, and after DCB are rotated the correct order has 8988. been achieved. 8989. 8990. However, Simple Simon, the train driver, is not known for his efficiency, 8991. so he always solves the problem by initially getting carriage A in the 8992. correct place, then carriage B, and so on. 8993. 8994. Using four carriages, the worst possible arrangements for Simon, which we 8995. shall call maximix arrangements, are DACB and DBAC; each requiring him 8996. five rotations (although, using the most efficient approach, they could be 8997. solved using just three rotations). The process he uses for DACB is shown 8998. below. 8999. 9000. It can be verified that there are 24 maximix arrangements for six 9001. carriages, of which the tenth lexicographic maximix arrangement is DFAECB. 9002. 9003. Find the 2011^th lexicographic maximix arrangement for eleven carriages. 9004. 9005. 9006. p_336_maximix.gif 9007. Answer: 7968e48fc692ce25bf7f5494f4ab6814 9008. 9009. 9010. Problem 337 9011. =========== 9012. 9013. 9014. Let {a, a,..., a[n]} be an integer sequence of length n such that: 9015. 9016. • a = 6 9017. • for all 1 ≤ i < n : φ(a[i]) < φ(a[i+1]) < a[i] < a[i+1] ^1 9018. 9019. Let S(N) be the number of such sequences with a[n] ≤ N. 9020. For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}. 9021. We can verify that S(100) = 482073668 and S(10 000) mod 10^8 = 73808307. 9022. 9023. Find S(20 000 000) mod 10^8. 9024. 9025. ^1 φ denotes Euler's totient function. 9026. 9027. 9028. Answer: 85068035 9029. 9030. 9031. Problem 338 9032. =========== 9033. 9034. 9035. A rectangular sheet of grid paper with integer dimensions w × h is given. 9036. Its grid spacing is 1. 9037. When we cut the sheet along the grid lines into two pieces and rearrange 9038. those pieces without overlap, we can make new rectangles with different 9039. dimensions. 9040. 9041. For example, from a sheet with dimensions 9 × 4 , we can make rectangles 9042. with dimensions 18 × 2, 12 × 3 and 6 × 6 by cutting and rearranging as 9043. below: 9044. 9045. Similarly, from a sheet with dimensions 9 × 8 , we can make rectangles 9046. with dimensions 18 × 4 and 12 × 6 . 9047. 9048. For a pair w and h, let F(w,h) be the number of distinct rectangles that 9049. can be made from a sheet with dimensions w × h . 9050. For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2. 9051. Note that rectangles congruent to the initial one are not counted in 9052. F(w,h). 9053. Note also that rectangles with dimensions w × h and dimensions h × w are 9054. not considered distinct. 9055. 9056. For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h 9057. which satisfy 0 < h ≤ w ≤ N. 9058. We can verify that G(10) = 55, G(10^3) = 971745 and G(10^5) = 9992617687. 9059. 9060. Find G(10^12). Give your answer modulo 10^8. 9061. 9062. 9063. p_338_gridpaper.gif 9064. Answer: 15614292 9065. 9066. 9067. Problem 339 9068. =========== 9069. 9070. 9071. "And he came towards a valley, through which ran a river; and the borders 9072. of the valley were wooded, and on each side of the river were level 9073. meadows. And on one side of the river he saw a flock of white sheep, and 9074. on the other a flock of black sheep. And whenever one of the white sheep 9075. bleated, one of the black sheep would cross over and become white; and 9076. when one of the black sheep bleated, one of the white sheep would cross 9077. over and become black." 9078. en.wikisource.org 9079. 9080. Initially each flock consists of n sheep. Each sheep (regardless of 9081. colour) is equally likely to be the next sheep to bleat. After a sheep has 9082. bleated and a sheep from the other flock has crossed over, Peredur may 9083. remove a number of white sheep in order to maximize the expected final 9084. number of black sheep. Let E(n) be the expected final number of black 9085. sheep if Peredur uses an optimal strategy. 9086. 9087. You are given that E(5) = 6.871346 rounded to 6 places behind the decimal 9088. point. 9089. Find E(10 000) and give your answer rounded to 6 places behind the decimal 9090. point. 9091. 9092. 9093. Visible links 9094. 1. http://en.wikisource.org/wiki/The_Mabinogion/Peredur_the_Son_of_Evrawc 9095. Answer: 0be02210b2d2212d37d026478093c457 9096. 9097. 9098. Problem 340 9099. =========== 9100. 9101. 9102. For fixed integers a, b, c, define the crazy function F(n) as follows: 9103. F(n) = n - c for all n > b 9104. F(n) = F(a + F(a + F(a + F(a + n)))) for all n ≤ b. 9105. 9106. Also, define S(a, b, c) = . 9107. 9108. For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000) 9109. = 2040. 9110. Also, S(50, 2000, 40) = 5204240. 9111. 9112. Find the last 9 digits of S(21^7, 7^21, 12^7). 9113. 9114. 9115. p_340_formula.gif 9116. Answer: fc838afe9ecde39bbe230923d7b50775 9117. 9118. 9119. Problem 341 9120. =========== 9121. 9122. 9123. The Golomb's self-describing sequence {G(n)} is the only nondecreasing 9124. sequence of natural numbers such that n appears exactly G(n) times in the 9125. sequence. The values of G(n) for the first few n are 9126. 9127. n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 … 9128. G(n) 1 2 2 3 3 4 4 4 5 5 5 6 6 6 6 … 9129. 9130. You are given that G(10^3) = 86, G(10^6) = 6137. 9131. You are also given that ΣG(n^3) = 153506976 for 1 ≤ n < 10^3. 9132. 9133. Find ΣG(n^3) for 1 ≤ n < 10^6. 9134. 9135. 9136. Answer: 7c163c3b4886943667b5c89db0a6cd02 9137. 9138. 9139. Problem 342 9140. =========== 9141. 9142. 9143. Consider the number 50. 9144. 50^2 = 2500 = 2^2 × 5^4, so φ(2500) = 2 × 4 × 5^3 = 8 × 5^3 = 2^3 × 5^3. 9145. ^1 9146. So 2500 is a square and φ(2500) is a cube. 9147. 9148. Find the sum of all numbers n, 1 < n < 10^10 such that φ(n^2) is a cube. 9149. 9150. ^1 φ denotes Euler's totient function. 9151. 9152. 9153. Answer: 0e9add0383d4116c7c5cb3dc73fc0536 9154. 9155. 9156. Problem 343 9157. =========== 9158. 9159. 9160. For any positive integer k, a finite sequence a[i] of fractions x[i]/y[i] 9161. is defined by: 9162. a = 1/k and 9163. a[i] = (x[i-1]+1)/(y[i-1]-1) reduced to lowest terms for i>1. 9164. When a[i] reaches some integer n, the sequence stops. (That is, when 9165. y[i]=1.) 9166. Define f(k) = n. 9167. For example, for k = 20: 9168. 9169. 1/20 → 2/19 → 3/18 = 1/6 → 2/5 → 3/4 → 4/3 → 5/2 → 6/1 = 6 9170. 9171. So f(20) = 6. 9172. 9173. Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k^3) = 118937 for 1 ≤ k ≤ 100. 9174. 9175. Find Σf(k^3) for 1 ≤ k ≤ 2×10^6. 9176. 9177. 9178. Answer: 0e10bd111425ad8e1343ac79dac7bb0e 9179. 9180. 9181. Problem 344 9182. =========== 9183. 9184. 9185. One variant of N.G. de Bruijn's silver dollar game can be described as 9186. follows: 9187. 9188. On a strip of squares a number of coins are placed, at most one coin per 9189. square. Only one coin, called the silver dollar, has any value. Two 9190. players take turns making moves. At each turn a player must make either a 9191. regular or a special move. 9192. 9193. A regular move consists of selecting one coin and moving it one or more 9194. squares to the left. The coin cannot move out of the strip or jump on or 9195. over another coin. 9196. 9197. Alternatively, the player can choose to make the special move of pocketing 9198. the leftmost coin rather than making a regular move. If no regular moves 9199. are possible, the player is forced to pocket the leftmost coin. 9200. 9201. The winner is the player who pockets the silver dollar. 9202. 9203. A winning configuration is an arrangement of coins on the strip where the 9204. first player can force a win no matter what the second player does. 9205. 9206. Let W(n,c) be the number of winning configurations for a strip of n 9207. squares, c worthless coins and one silver dollar. 9208. 9209. You are given that W(10,2) = 324 and W(100,10) = 1514704946113500. 9210. 9211. Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003 9212. · 1 000 033). 9213. 9214. 9215. p_344_silverdollar.gif 9216. Answer: 38e7b980b38fcac89b3e267e328cd292 9217. 9218. 9219. Problem 345 9220. =========== 9221. 9222. 9223. We define the Matrix Sum of a matrix as the maximum sum of matrix elements 9224. with each element being the only one in his row and column. For example, 9225. the Matrix Sum of the matrix below equals 3315 ( = 863 + 383 + 343 + 959 + 9226. 767): 9227. 9228. 7 53 183 439 863 9229. 497 383 563 79 973 9230. 287 63 343 169 583 9231. 627 343 773 959 943 9232. 767 473 103 699 303 9233. 9234. Find the Matrix Sum of: 9235. 9236. 7 53 183 439 863 497 383 563 79 973 287 63 343 169 583 9237. 627 343 773 959 943 767 473 103 699 303 957 703 583 639 913 9238. 447 283 463 29 23 487 463 993 119 883 327 493 423 159 743 9239. 217 623 3 399 853 407 103 983 89 463 290 516 212 462 350 9240. 960 376 682 962 300 780 486 502 912 800 250 346 172 812 350 9241. 870 456 192 162 593 473 915 45 989 873 823 965 425 329 803 9242. 973 965 905 919 133 673 665 235 509 613 673 815 165 992 326 9243. 322 148 972 962 286 255 941 541 265 323 925 281 601 95 973 9244. 445 721 11 525 473 65 511 164 138 672 18 428 154 448 848 9245. 414 456 310 312 798 104 566 520 302 248 694 976 430 392 198 9246. 184 829 373 181 631 101 969 613 840 740 778 458 284 760 390 9247. 821 461 843 513 17 901 711 993 293 157 274 94 192 156 574 9248. 34 124 4 878 450 476 712 914 838 669 875 299 823 329 699 9249. 815 559 813 459 522 788 168 586 966 232 308 833 251 631 107 9250. 813 883 451 509 615 77 281 613 459 205 380 274 302 35 805 9251. 9252. 9253. Answer: 13938 9254. 9255. 9256. Problem 346 9257. =========== 9258. 9259. 9260. The number 7 is special, because 7 is 111 written in base 2, and 11 9261. written in base 6 9262. (i.e. 7 = 11 = 111). In other words, 7 is a repunit in at least 9263. two bases b > 1. 9264. 9265. We shall call a positive integer with this property a strong repunit. It 9266. can be verified that there are 8 strong repunits below 50: 9267. {1,7,13,15,21,31,40,43}. 9268. Furthermore, the sum of all strong repunits below 1000 equals 15864. 9269. 9270. Find the sum of all strong repunits below 10^12. 9271. 9272. Answer: a17874b5a9ec9d7fc8c6489ab8ff29b9 9273. 9274. 9275. Problem 347 9276. =========== 9277. 9278. 9279. The largest integer ≤ 100 that is only divisible by both the primes 2 and 9280. 3 is 96, as 96=32*3=2^5*3.For two distinct primes p and q let M(p,q,N) be 9281. the largest positive integer ≤N only divisibleby both p and q and 9282. M(p,q,N)=0 if such a positive integer does not exist. 9283. 9284. E.g. M(2,3,100)=96. 9285. M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. 9286. Also M(2,73,100)=0 because there does not exist a positive integer ≤ 100 9287. that is divisible by both 2 and 73. 9288. 9289. Let S(N) be the sum of all distinct M(p,q,N).S(100)=2262. 9290. 9291. Find S(10 000 000). 9292. 9293. 9294. Answer: 96ce0eabcbe7a2b2eb1197a1bcc5d37b 9295. 9296. 9297. Problem 348 9298. =========== 9299. 9300. 9301. Many numbers can be expressed as the sum of a square and a cube. Some of 9302. them in more than one way. 9303. 9304. Consider the palindromic numbers that can be expressed as the sum of a 9305. square and a cube, both greater than 1, in exactly 4 different ways. 9306. For example, 5229225 is a palindromic number and it can be expressed in 9307. exactly 4 different ways: 9308. 9309. 2285^2 + 20^3 9310. 2223^2 + 66^3 9311. 1810^2 + 125^3 9312. 1197^2 + 156^3 9313. 9314. Find the sum of the five smallest such palindromic numbers. 9315. 9316. 9317. Answer: f286f9159fc20aeb97a8bf8396ba64de 9318. 9319. 9320. Problem 349 9321. =========== 9322. 9323. 9324. An ant moves on a regular grid of squares that are coloured either black 9325. or white. 9326. The ant is always oriented in one of the cardinal directions (left, right, 9327. up or down) and moves from square to adjacent square according to the 9328. following rules: 9329. - if it is on a black square, it flips the color of the square to white, 9330. rotates 90 degrees counterclockwise and moves forward one square. 9331. - if it is on a white square, it flips the color of the square to black, 9332. rotates 90 degrees clockwise and moves forward one square. 9333. 9334. Starting with a grid that is entirely white, how many squares are black 9335. after 10^18 moves of the ant? 9336. 9337. 9338. Answer: 412b0faec10b3adb415363d2df26530d 9339. 9340. 9341. Problem 350 9342. =========== 9343. 9344. 9345. A list of size n is a sequence of n natural numbers. 9346. Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11). 9347. 9348. The greatest common divisor, or gcd, of a list is the largest natural 9349. number that divides all entries of the list. 9350. Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11. 9351. 9352. The least common multiple, or lcm, of a list is the smallest natural 9353. number divisible by each entry of the list. 9354. Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11. 9355. 9356. Let f(G, L, N) be the number of lists of size N with gcd ≥ G and lcm ≤ L. 9357. For example: 9358. 9359. f(10, 100, 1) = 91. 9360. f(10, 100, 2) = 327. 9361. f(10, 100, 3) = 1135. 9362. f(10, 100, 1000) mod 101^4 = 3286053. 9363. 9364. Find f(10^6, 10^12, 10^18) mod 101^4. 9365. 9366. 9367. Answer: 84664213 9368. 9369. 9370. Problem 351 9371. =========== 9372. 9373. 9374. A hexagonal orchard of order n is a triangular lattice made up of points 9375. within a regular hexagon with side n. The following is an example of a 9376. hexagonal orchard of order 5: 9377. 9378. Highlighted in green are the points which are hidden from the center by a 9379. point closer to it. It can be seen that for a hexagonal orchard of order 9380. 5, 30 points are hidden from the center. 9381. 9382. Let H(n) be the number of points hidden from the center in a hexagonal 9383. orchard of order n. 9384. 9385. H(5) = 30. H(10) = 138. H(1 000) = 1177848. 9386. 9387. Find H(100 000 000). 9388. 9389. 9390. p_351_hexorchard.png 9391. Answer: 338481092e945257756075a8d03978fd 9392. 9393. 9394. Problem 352 9395. =========== 9396. 9397. 9398. Each one of the 25 sheep in a flock must be tested for a rare virus, known 9399. to affect 2% of the sheep population.An accurate and extremely sensitive 9400. PCR test exists for blood samples, producing a clear positive / negative 9401. result, but it is very time-consuming and expensive. 9402. 9403. Because of the high cost, the vet-in-charge suggests that instead of 9404. performing 25 separate tests, the following procedure can be used instead: 9405. 9406. The sheep are split into 5 groups of 5 sheep in each group. For each 9407. group, the 5 samples are mixed together and a single test is performed. 9408. Then, 9409. 9410. • If the result is negative, all the sheep in that group are deemed to 9411. be virus-free. 9412. • If the result is positive, 5 additional tests will be performed (a 9413. separate test for each animal) to determine the affected 9414. individual(s). 9415. 9416. Since the probability of infection for any specific animal is only 0.02, 9417. the first test (on the pooled samples) for each group will be: 9418. 9419. • Negative (and no more tests needed) with probability 0.98^5 = 9420. 0.9039207968. 9421. • Positive (5 additional tests needed) with probability 1 - 0.9039207968 9422. = 0.0960792032. 9423. 9424. Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5 9425. = 1.480396016. 9426. Consequently, all 5 groups can be screened using an average of only 9427. 1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of more 9428. than 70% ! 9429. 9430. Although the scheme we have just described seems to be very efficient, it 9431. can still be improved considerably (always assuming that the test is 9432. sufficiently sensitive and that there are no adverse effects caused by 9433. mixing different samples). E.g.: 9434. 9435. • We may start by running a test on a mixture of all the 25 samples. It 9436. can be verified that in about 60.35% of the cases this test will be 9437. negative, thus no more tests will be needed. Further testing will only 9438. be required for the remaining 39.65% of the cases. 9439. • If we know that at least one animal in a group of 5 is infected and 9440. the first 4 individual tests come out negative, there is no need to 9441. run a test on the fifth animal (we know that it must be infected). 9442. • We can try a different number of groups / different number of animals 9443. in each group, adjusting those numbers at each level so that the total 9444. expected number of tests will be minimised. 9445. 9446. To simplify the very wide range of possibilities, there is one restriction 9447. we place when devising the most cost-efficient testing scheme: whenever we 9448. start with a mixed sample, all the sheep contributing to that sample must 9449. be fully screened (i.e. a verdict of infected / virus-free must be reached 9450. for all of them) before we start examining any other animals. 9451. 9452. For the current example, it turns out that the most cost-efficient testing 9453. scheme (we'll call it the optimal strategy) requires an average of just 9454. 4.155452 tests! 9455. 9456. Using the optimal strategy, let T(s,p) represent the average number of 9457. tests needed to screen a flock of s sheep for a virus having probability p 9458. to be present in any individual. 9459. Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25, 9460. 0.10) = 12.702124. 9461. 9462. Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50. 9463. Give your answer rounded to six decimal places. 9464. 9465. 9466. Answer: 2e74b2fb574d6318cdbf2a41ad006de7 9467. 9468. 9469. Problem 353 9470. =========== 9471. 9472. 9473. A moon could be described by the sphere C(r) with centre (0,0,0) and 9474. radius r. 9475. 9476. There are stations on the moon at the points on the surface of C(r) with 9477. integer coordinates. The station at (0,0,r) is called North Pole station, 9478. the station at (0,0,-r) is called South Pole station. 9479. 9480. All stations are connected with each other via the shortest road on the 9481. great arc through the stations. A journey between two stations is risky. 9482. If d is the length of the road between two stations, (d/(π r))^2 is a 9483. measure for the risk of the journey (let us call it the risk of the road). 9484. If the journey includes more than two stations, the risk of the journey is 9485. the sum of risks of the used roads. 9486. 9487. A direct journey from the North Pole station to the South Pole station has 9488. the length πr and risk 1. The journey from the North Pole station to the 9489. South Pole station via (0,r,0) has the same length, but a smaller risk: 9490. (½πr/(πr))^2+(½πr/(πr))^2=0.5. 9491. 9492. The minimal risk of a journey from the North Pole station to the South 9493. Pole station on C(r) is M(r). 9494. 9495. You are given that M(7)=0.1784943998 rounded to 10 digits behind the 9496. decimal point. 9497. 9498. Find ∑M(2^n-1) for 1≤n≤15. 9499. 9500. Give your answer rounded to 10 digits behind the decimal point in the form 9501. a.bcdefghijk. 9502. 9503. 9504. Answer: 211b5626459be71baefc78478d18bdc3 9505. 9506. 9507. Problem 354 9508. =========== 9509. 9510. 9511. Consider a honey bee's honeycomb where each cell is a perfect regular 9512. hexagon with side length 1. 9513. 9514. One particular cell is occupied by the queen bee. 9515. For a positive real number L, let B(L) count the cells with distance L 9516. from the queen bee cell (all distances are measured from centre to 9517. centre); you may assume that the honeycomb is large enough to accommodate 9518. for any distance we wish to consider. 9519. For example, B(√3) = 6, B(√21) = 12 and B(111 111 111) = 54. 9520. 9521. Find the number of L ≤ 5·10^11 such that B(L) = 450. 9522. 9523. 9524. p_354_bee_honeycomb.png 9525. Answer: 58065134 9526. 9527. 9528. Problem 355 9529. =========== 9530. 9531. 9532. Define Co(n) to be the maximal possible sum of a set of mutually co-prime 9533. elements from {1, 2, ..., n}. 9534. For example Co(10) is 30 and hits that maximum on the subset 9535. {1, 5, 7, 8, 9}. 9536. 9537. You are given that Co(30) = 193 and Co(100) = 1356. 9538. 9539. Find Co(200000). 9540. 9541. 9542. Answer: 41cb97b6d02878d79f8b2e3b6c74920a 9543. 9544. 9545. Problem 356 9546. =========== 9547. 9548. 9549. Let a[n] be the largest real root of a polynomial g(x) = x^3 - 2^n·x^2 + 9550. n. 9551. For example, a = 3.86619826... 9552. 9553. Find the last eight digits of. 9554. 9555. Note: represents the floor function. 9556. 9557. 9558. p_356_cubicpoly1.gif 9559. p_356_cubicpoly2.gif 9560. Answer: 28010159 9561. 9562. 9563. Problem 357 9564. =========== 9565. 9566. 9567. Consider the divisors of 30: 1,2,3,5,6,10,15,30. 9568. It can be seen that for every divisor d of 30, d+30/d is prime. 9569. 9570. Find the sum of all positive integers n not exceeding 100 000 000 9571. such thatfor every divisor d of n, d+n/d is prime. 9572. 9573. 9574. Answer: ed25b13b18a21c1077fed00ef42f503b 9575. 9576. 9577. Problem 358 9578. =========== 9579. 9580. 9581. A cyclic number with n digits has a very interesting property: 9582. When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly 9583. the same digits, in the same order, but rotated in a circular fashion! 9584. 9585. The smallest cyclic number is the 6-digit number 142857 : 9586. 142857 × 1 = 142857 9587. 142857 × 2 = 285714 9588. 142857 × 3 = 428571 9589. 142857 × 4 = 571428 9590. 142857 × 5 = 714285 9591. 142857 × 6 = 857142 9592. 9593. The next cyclic number is 0588235294117647 with 16 digits : 9594. 0588235294117647 × 1 = 0588235294117647 9595. 0588235294117647 × 2 = 1176470588235294 9596. 0588235294117647 × 3 = 1764705882352941 9597. ... 9598. 0588235294117647 × 16 = 9411764705882352 9599. 9600. Note that for cyclic numbers, leading zeros are important. 9601. 9602. There is only one cyclic number for which, the eleven leftmost digits are 9603. 00000000137 and the five rightmost digits are 56789 (i.e., it has the form 9604. 00000000137...56789 with an unknown number of digits in the middle). Find 9605. the sum of all its digits. 9606. 9607. 9608. Answer: 359e1ec8aeaa3932b54f2a5d20fa4f73 9609. 9610. 9611. Problem 359 9612. =========== 9613. 9614. 9615. An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get 9616. a room at Hilbert's newest infinite hotel. The hotel contains an infinite 9617. number of floors (numbered 1, 2, 3, etc.), and each floor contains an 9618. infinite number of rooms (numbered 1, 2, 3, etc.). 9619. 9620. Initially the hotel is empty. Hilbert declares a rule on how the n^th 9621. person is assigned a room: person n gets the first vacant room in the 9622. lowest numbered floor satisfying either of the following: 9623. 9624. • the floor is empty 9625. • the floor is not empty, and if the latest person taking a room in that 9626. floor is person m, then m + n is a perfect square 9627. 9628. Person 1 gets room 1 in floor 1 since floor 1 is empty. 9629. Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect 9630. square. 9631. Person 2 instead gets room 1 in floor 2 since floor 2 is empty. 9632. Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square. 9633. 9634. Eventually, every person in the line gets a room in the hotel. 9635. 9636. Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no 9637. person occupies the room. Here are a few examples: 9638. P(1, 1) = 1 9639. P(1, 2) = 3 9640. P(2, 1) = 2 9641. P(10, 20) = 440 9642. P(25, 75) = 4863 9643. P(99, 100) = 19454 9644. 9645. Find the sum of all P(f, r) for all positive f and r such that f × r = 9646. 71328803586048 and give the last 8 digits as your answer. 9647. 9648. 9649. Answer: 40632119 9650. 9651. 9652. Problem 360 9653. =========== 9654. 9655. 9656. Given two points (x,y,z) and (x,y,z) in three 9657. dimensional space, the Manhattan distance between those points is defined 9658. as 9659. |x-x|+|y-y|+|z-z|. 9660. 9661. Let C(r) be a sphere with radius r and center in the origin O(0,0,0). 9662. Let I(r) be the set of all points with integer coordinates on the surface 9663. of C(r). 9664. Let S(r) be the sum of the Manhattan distances of all elements of I(r) to 9665. the origin O. 9666. 9667. E.g. S(45)=34518. 9668. 9669. Find S(10^10). 9670. 9671. 9672. Answer: 82ec91527315eafb7e3acc139eeeb8eb 9673. 9674. 9675. Problem 361 9676. =========== 9677. 9678. 9679. The Thue-Morse sequence {T[n]} is a binary sequence satisfying: 9680. 9681. • T = 0 9682. • T[2n] = T[n] 9683. • T[2n+1] = 1 - T[n] 9684. 9685. The first several terms of {T[n]} are given as follows: 9686. 01101001100101101001011001101001.... 9687. 9688. We define {A[n]} as the sorted sequence of integers such that the binary 9689. expression of each element appears as a subsequence in {T[n]}. 9690. For example, the decimal number 18 is expressed as 10010 in binary. 10010 9691. appears in {T[n]} (T to T), so 18 is an element of {A[n]}. 9692. The decimal number 14 is expressed as 1110 in binary. 1110 never appears 9693. in {T[n]}, so 14 is not an element of {A[n]}. 9694. 9695. The first several terms of A[n] are given as follows: 9696. 9697. n 0 1 2 3 4 5 6 7 8 9 10 11 12 … 9698. A[n] 0 1 2 3 4 5 6 9 10 11 12 13 18 … 9699. 9700. We can also verify that A = 3251 and A = 80852364498. 9701. 9702. Find the last 9 digits of . 9703. 9704. 9705. p_361_Thue-Morse1.gif 9706. Answer: 178476944 9707. 9708. 9709. Problem 362 9710. =========== 9711. 9712. 9713. Consider the number 54. 9714. 54 can be factored in 7 distinct ways into one or more factors larger than 9715. 1: 9716. 54, 2×27, 3×18, 6×9, 3×3×6, 2×3×9 and 2×3×3×3. 9717. If we require that the factors are all squarefree only two ways remain: 9718. 3×3×6 and 2×3×3×3. 9719. 9720. Let's call Fsf(n) the number of ways n can be factored into one or more 9721. squarefree factors larger than 1, soFsf(54)=2. 9722. 9723. Let S(n) be ∑Fsf(k) for k=2 to n. 9724. 9725. S(100)=193. 9726. 9727. Find S(10 000 000 000). 9728. 9729. 9730. Answer: b62f0d524bec8653ba7b8a2cab70260b 9731. 9732. 9733. Problem 363 9734. =========== 9735. 9736. A cubic Bézier curve is defined by four points: P, P, P and P. 9737. 9738. The curve is constructed as follows: 9739. On the segments PP, PP and PP the points Q,Q and 9740. Q are drawn such that 9741. PQ/PP=PQ/PP=PQ/PP=t (t in [0,1]). 9742. On the segments QQ and QQ the points R and R are drawn 9743. such thatQR/QQ=QR/QQ=t for the same value of t. 9744. On the segment RR the point B is drawn such that RB/RR=t 9745. for the same value of t.The Bézier curve defined by the points P, P, 9746. P, P is the locus of B as Q takes all possible positions on the 9747. segment PP. (Please note that for all points the value of t is the 9748. same.) 9749. 9750. Applet 9751. 9752. In the applet to the right you can drag the points P, P, P and 9753. P to see what the Bézier curve (green curve) defined by those points 9754. looks like. You can also drag the point Q along the segment PP. 9755. 9756. From the construction it is clear that the Bézier curve will be tangent to 9757. the segments PP in P and PP in P. 9758. 9759. A cubic Bézier curve with P=(1,0), P=(1,v), P=(v,1) and 9760. P=(0,1) is used to approximate a quarter circle. 9761. The value v>0 is chosen such that the area enclosed by the lines OP, 9762. OP and the curve is equal to ^π/ (the area of the quarter circle). 9763. 9764. By how many percent does the length of the curve differ from the length of 9765. the quarter circle? 9766. That is, if L is the length of the curve, calculate 100*^(L-π/2)/[(π/2)]. 9767. Give your answer rounded to 10 digits behind the decimal point. 9768. 9769. 9770. Visible links 9771. 1. CabriJava.class 9772. Answer: 2bc63386b7cccc64c67f90e719936143 9773. 9774. 9775. Problem 364 9776. =========== 9777. 9778. 9779. There are N seats in a row. N people come after each other to fill the 9780. seats according to the following rules: 9781. 9782. 1. If there is any seat whose adjacent seat(s) are not occupied take such 9783. a seat. 9784. 2. If there is no such seat and there is any seat for which only one 9785. adjacent seat is occupied take such a seat. 9786. 3. Otherwise take one of the remaining available seats. 9787. 9788. Let T(N) be the number of possibilities that N seats are occupied by N 9789. people with the given rules. 9790. The following figure shows T(4)=8. 9791. 9792. We can verify that T(10) = 61632 and T(1 000) mod 100 000 007 = 47255094. 9793. 9794. Find T(1 000 000) mod 100 000 007. 9795. 9796. 9797. p_364_comf_dist.gif 9798. Answer: 44855254 9799. 9800. 9801. Problem 365 9802. =========== 9803. 9804. 9805. The binomial coeffient C(10^18,10^9) is a number with more than 9 billion 9806. (9×10^9) digits. 9807. 9808. Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m. 9809. 9810. Calculate ∑M(10^18,10^9,p*q*r) for 1000<p<q<r<5000 and p,q,r prime. 9811. 9812. 9813. Answer: 53addf69042b0cefbeb94f3bd3224918 9814. 9815. 9816. Problem 366 9817. =========== 9818. 9819. 9820. Two players, Anton and Bernhard, are playing the following game. 9821. There is one pile of n stones. 9822. The first player may remove any positive number of stones, but not the 9823. whole pile. 9824. Thereafter, each player may remove at most twice the number of stones his 9825. opponent took on the previous move. 9826. The player who removes the last stone wins. 9827. 9828. E.g. n=5 9829. If the first player takes anything more than one stone the next player 9830. will be able to take all remaining stones. 9831. If the first player takes one stone, leaving four, his opponent will take 9832. also one stone, leaving three stones. 9833. The first player cannot take all three because he may take at most 2x1=2 9834. stones. So let's say he takes also one stone, leaving 2. The second player 9835. can take the two remaining stones and wins. 9836. So 5 is a losing position for the first player. 9837. For some winning positions there is more than one possible move for the 9838. first player. 9839. E.g. when n=17 the first player can remove one or four stones. 9840. 9841. Let M(n) be the maximum number of stones the first player can take from a 9842. winning position at his first turn and M(n)=0 for any other position. 9843. 9844. ∑M(n) for n≤100 is 728. 9845. 9846. Find ∑M(n) for n≤10^18.Give your answer modulo 10^8. 9847. 9848. 9849. Answer: 88351299 9850. 9851. 9852. Problem 367 9853. =========== 9854. 9855. 9856. Bozo sort, not to be confused with the slightly less efficient bogo sort, 9857. consists out of checking if the input sequence is sorted and if not 9858. swapping randomly two elements. This is repeated until eventually the 9859. sequence is sorted. 9860. 9861. If we consider all permutations of the first 4 natural numbers as input 9862. the expectation value of the number of swaps, averaged over all 4! input 9863. sequences is 24.75. 9864. The already sorted sequence takes 0 steps. 9865. 9866. In this problem we consider the following variant on bozo sort. 9867. If the sequence is not in order we pick three elements at random and 9868. shuffle these three elements randomly. 9869. All 3!=6 permutations of those three elements are equally likely. 9870. The already sorted sequence will take 0 steps. 9871. If we consider all permutations of the first 4 natural numbers as input 9872. the expectation value of the number of shuffles, averaged over all 4! 9873. input sequences is 27.5. 9874. Consider as input sequences the permutations of the first 11 natural 9875. numbers. 9876. Averaged over all 11! input sequences, what is the expected number of 9877. shuffles this sorting algorithm will perform? 9878. 9879. Give your answer rounded to the nearest integer. 9880. 9881. 9882. Answer: 48271207 9883. 9884. 9885. Problem 368 9886. =========== 9887. 9888. 9889. The harmonic series 1 + 1 + 1 + 1 + ... is well known to be divergent. 9890. 2 3 4 9891. 9892. If we however omit from this series every term where the denominator has a 9893. 9 in it, the series remarkably enough converges to approximately 9894. 22.9206766193. 9895. This modified harmonic series is called the Kempner series. 9896. 9897. Let us now consider another modified harmonic series by omitting from the 9898. harmonic series every term where the denominator has 3 or more equal 9899. consecutive digits.One can verify that out of the first 1200 terms of the 9900. harmonic series, only 20 terms will be omitted. 9901. These 20 omitted terms are: 9902. 9903. 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 9904. 111 222 333 444 555 666 777 888 999 1000 1110 9905. 9906. 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 and 1 . 9907. 1111 1112 1113 1114 1115 1116 1117 1118 1119 9908. 9909. This series converges as well. 9910. 9911. Find the value the series converges to. 9912. Give your answer rounded to 10 digits behind the decimal point. 9913. 9914. 9915. Answer: bfb15c388f4721cbd5eb89f17be2eef2 9916. 9917. 9918. Problem 369 9919. =========== 9920. 9921. 9922. In a standard 52 card deck of playing cards, a set of 4 cards is a Badugi 9923. if it contains 4 cards with no pairs and no two cards of the same suit. 9924. 9925. Let f(n) be the number of ways to choose n cards with a 4 card subset that 9926. is a Badugi. For example, there are 2598960 ways to choose five cards from 9927. a standard 52 card deck, of which 514800 contain a 4 card subset that is a 9928. Badugi, so f(5) = 514800. 9929. 9930. Find ∑f(n) for 4 ≤ n ≤ 13. 9931. 9932. 9933. Answer: 0f8828f58dbac4f15f296c79b686ed0e 9934. 9935. 9936. Problem 370 9937. =========== 9938. 9939. 9940. Let us define a geometric triangle as an integer sided triangle with sides 9941. a ≤ b ≤ c so that its sides form a geometric progression, i.e. 9942. b^2 = a · c . 9943. 9944. An example of such a geometric triangle is the triangle with sides a = 9945. 144, b = 156 and c = 169. 9946. 9947. There are 861805 geometric triangles with perimeter ≤ 10^6 . 9948. 9949. How many geometric triangles exist with perimeter ≤ 2.5·10^13 ? 9950. 9951. 9952. Answer: 85b5048e25677205555a5308991c2e04 9953. 9954. 9955. Problem 371 9956. =========== 9957. 9958. 9959. Oregon licence plates consist of three letters followed by a three digit 9960. number (each digit can be from [0..9]). 9961. While driving to work Seth plays the following game: 9962. Whenever the numbers of two licence plates seen on his trip add to 1000 9963. that's a win. 9964. 9965. E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as 9966. he sees them in the same trip). 9967. 9968. Find the expected number of plates he needs to see for a win. 9969. Give your answer rounded to 8 decimal places behind the decimal point. 9970. 9971. Note: We assume that each licence plate seen is equally likely to have any 9972. three digit number on it. 9973. 9974. 9975. Answer: 537403a97924621c604ce5ab6288b97d 9976. 9977. 9978. Problem 372 9979. =========== 9980. 9981. 9982. Let R(M, N) be the number of lattice points (x, y) which satisfy M<x≤N, 9983. M<y≤N and is odd. 9984. We can verify that R(0, 100) = 3019 and R(100, 10000) = 29750422. 9985. Find R(2·10^6, 10^9). 9986. 9987. Note: represents the floor function. 9988. 9989. 9990. p_372_pencilray1.jpg 9991. p_372_pencilray2.gif 9992. Answer: 5fdeda0dca23d12ae3eb1763b2c6f5ea 9993. 9994. 9995. Problem 373 9996. =========== 9997. 9998. 9999. Every triangle has a circumscribed circle that goes through the three 10000. vertices.Consider all integer sided triangles for which the radius of the 10001. circumscribed circle is integral as well. 10002. 10003. Let S(n) be the sum of the radii of the circumscribed circles of all such 10004. triangles for which the radius does not exceed n. 10005. 10006. S(100)=4950 and S(1200)=1653605. 10007. 10008. Find S(10^7). 10009. 10010. 10011. Answer: 888d60a6b2b4b9146d7c9c14ffd82673 10012. 10013. 10014. Problem 374 10015. =========== 10016. 10017. 10018. An integer partition of a number n is a way of writing n as a sum of 10019. positive integers. 10020. 10021. Partitions that differ only in the order of their summands are considered 10022. the same.A partition of n into distinct parts is a partition of n in which 10023. every part occurs at most once. 10024. 10025. The partitions of 5 into distinct parts are: 10026. 5, 4+1 and 3+2. 10027. 10028. Let f(n) be the maximum product of the parts of any such partition of n 10029. into distinct parts and let m(n) be the number of elements of any such 10030. partition of n with that product. 10031. 10032. So f(5)=6 and m(5)=2. 10033. 10034. For n=10 the partition with the largest product is 10=2+3+5, which gives 10035. f(10)=30 and m(10)=3. 10036. And their product, f(10)·m(10) = 30·3 = 90 10037. 10038. It can be verified that 10039. ∑f(n)·m(n) for 1 ≤ n ≤ 100 = 1683550844462. 10040. 10041. Find ∑f(n)·m(n) for 1 ≤ n ≤ 10^14. 10042. Give your answer modulo 982451653, the 50 millionth prime. 10043. 10044. 10045. Answer: 6fcb063062076b5aaaff3e3cd03e4b2f 10046. 10047. 10048. Problem 375 10049. =========== 10050. 10051. 10052. Let S[n] be an integer sequence produced with the following pseudo-random 10053. number generator: 10054. 10055. S =[ ] 290797[ ] 10056. S[n+1] =[ ] S[n]^2 mod 50515093 10057. 10058. Let A(i, j) be the minimum of the numbers S[i], S[i+1], ... , S[j] for i ≤ 10059. j. 10060. Let M(N) = ΣA(i, j) for 1 ≤ i ≤ j ≤ N. 10061. We can verify that M(10) = 432256955 and M(10 000) = 3264567774119. 10062. 10063. Find M(2 000 000 000). 10064. 10065. 10066. Answer: 68a12e3f2e4ccbae9c8555e547fbe096 10067. 10068. 10069. Problem 376 10070. =========== 10071. 10072. 10073. Consider the following set of dice with nonstandard pips: 10074. 10075. Die A: 1 4 4 4 4 4 10076. Die B: 2 2 2 5 5 5 10077. Die C: 3 3 3 3 3 6 10078. 10079. A game is played by two players picking a die in turn and rolling it. The 10080. player who rolls the highest value wins. 10081. 10082. If the first player picks die A and the second player picks die B we get 10083. P(second player wins) = 7/12 > 1/2 10084. 10085. If the first player picks die B and the second player picks die C we get 10086. P(second player wins) = 7/12 > 1/2 10087. 10088. If the first player picks die C and the second player picks die A we get 10089. P(second player wins) = 25/36 > 1/2 10090. 10091. So whatever die the first player picks, the second player can pick another 10092. die and have a larger than 50% chance of winning. 10093. A set of dice having this property is called a nontransitive set of dice. 10094. 10095. We wish to investigate how many sets of nontransitive dice exist. We will 10096. assume the following conditions: 10097. 10098. • There are three six-sided dice with each side having between 1 and N 10099. pips, inclusive. 10100. • Dice with the same set of pips are equal, regardless of which side on 10101. the die the pips are located. 10102. • The same pip value may appear on multiple dice; if both players roll 10103. the same value neither player wins. 10104. • The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set. 10105. 10106. For N = 7 we find there are 9780 such sets. 10107. How many are there for N = 30 ? 10108. 10109. 10110. Answer: c64df302990eb3738f8ec62ea6b66c0b 10111. 10112. 10113. Problem 377 10114. =========== 10115. 10116. 10117. There are 16 positive integers that do not have a zero in their digits and 10118. that have a digital sum equal to 5, namely: 10119. 5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111 10120. and 11111. 10121. Their sum is 17891. 10122. 10123. Let f(n) be the sum of all positive integers that do not have a zero in 10124. their digits and have a digital sum equal to n. 10125. 10126. Find . 10127. Give the last 9 digits as your answer. 10128. 10129. 10130. Answer: a915ccbae49de15208c88affba84d206 10131. 10132. 10133. Problem 378 10134. =========== 10135. 10136. 10137. Let T(n) be the n^th triangle number, so T(n) = n (n+1) . 10138. 2 10139. 10140. Let dT(n) be the number of divisors of T(n). 10141. E.g.:T(7) = 28 and dT(7) = 6. 10142. 10143. Let Tr(n) be the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n 10144. and dT(i) > dT(j) > dT(k). 10145. Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776. 10146. 10147. Find Tr(60 000 000). 10148. Give the last 18 digits of your answer. 10149. 10150. 10151. Answer: 336745dc9d90928596237c4b471a8927 10152. 10153. 10154. Problem 379 10155. =========== 10156. 10157. 10158. Let f(n) be the number of couples (x,y) with x and y positive integers, x 10159. ≤ y and the least common multiple of x and y equal to n. 10160. 10161. Let g be the summatory function of f, i.e.: g(n) = ∑ f(i) for 1 ≤ i ≤ n. 10162. 10163. You are given that g(10^6) = 37429395. 10164. 10165. Find g(10^12). 10166. 10167. 10168. Answer: de20f710cb6665c48795072197ad53e0 10169. 10170. 10171. Problem 380 10172. =========== 10173. 10174. 10175. An m×n maze is an m×n rectangular grid with walls placed between grid 10176. cells such that there is exactly one path from the top-left square to any 10177. other square. 10178. The following are examples of a 9×12 maze and a 15×20 maze: 10179. 10180. Let C(m,n) be the number of distinct m×n mazes. Mazes which can be formed 10181. by rotation and reflection from another maze are considered distinct. 10182. 10183. It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12) 10184. = 2.5720e46 (in scientific notation rounded to 5 significant digits). 10185. Find C(100,500) and write your answer in scientific notation rounded to 5 10186. significant digits. 10187. 10188. When giving your answer, use a lowercase e to separate mantissa and 10189. exponent.E.g. if the answer is 1234567891011 then the answer format would 10190. be 1.2346e12. 10191. 10192. 10193. Answer: c86d2f4c17c8134fbebed5d37a0f90d7 10194. 10195. 10196. Problem 381 10197. =========== 10198. 10199. 10200. For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5. 10201. 10202. For example, if p=7, 10203. (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 10204. 720+120+24+6+2 = 872. 10205. As 872 mod(7) = 4, S(7) = 4. 10206. 10207. It can be verified that ∑S(p) = 480 for 5 ≤ p < 100. 10208. 10209. Find ∑S(p) for 5 ≤ p < 10^8. 10210. 10211. 10212. Answer: 80c84973a9643e46d49d79d7284e7ff3 10213. 10214. 10215. Problem 382 10216. =========== 10217. 10218. 10219. A polygon is a flat shape consisting of straight line segments that are 10220. joined to form a closed chain or circuit. A polygon consists of at least 10221. three sides and does not self-intersect. 10222. 10223. A set S of positive numbers is said to generate a polygon P if: 10224. 10225. • no two sides of P are the same length, 10226. • the length of every side of P is in S, and 10227. • S contains no other value. 10228. 10229. For example: 10230. The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle). 10231. The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a 10232. quadrilateral). 10233. The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all. 10234. 10235. Consider the sequence s, defined as follows: 10236. 10237. • s = 1, s = 2, s = 3 10238. • s[n] = s[n-1] + s[n-3] for n > 3. 10239. 10240. Let U[n] be the set {s, s, ..., s[n]}. For example, U = {1, 2, 10241. 3, 4, 6, 9, 13, 19, 28, 41}. 10242. Let f(n) be the number of subsets of U[n] which generate at least one 10243. polygon. 10244. For example, f(5) = 7, f(10) = 501 and f(25) = 18635853. 10245. 10246. Find the last 9 digits of f(10^18). 10247. 10248. 10249. Answer: 56a121bcf3bb674d0d3ce561b6b24ea5 10250. 10251. 10252. Problem 383 10253. =========== 10254. 10255. 10256. Let f(n) be the largest integer x for which 5^x divides n. 10257. For example, f(625000) = 7. 10258. 10259. Let T(n) be the number of integers i which satisfy f((2·i-1)!) < 10260. 2·f(i!) and 1 ≤ i ≤ n. 10261. It can be verified that T(10^3) = 68 and T(10^9) = 2408210. 10262. 10263. Find T(10^18). 10264. 10265. 10266. Answer: c1bc7c945344e1967bfaced9ade895a0 10267. 10268. 10269. Problem 384 10270. =========== 10271. 10272. 10273. Define the sequence a(n) as the number of adjacent pairs of ones in the 10274. binary expansion of n (possibly overlapping). 10275. E.g.: a(5) = a(101) = 0, a(6) = a(110) = 1, a(7) = a(111) = 2 10276. 10277. Define the sequence b(n) = (-1)^a(n). 10278. This sequence is called the Rudin-Shapiro sequence. 10279. 10280. Also consider the summatory sequence of b(n): . 10281. 10282. The first couple of values of these sequences are: 10283. n 0 1 2 3 4 5 6 7 10284. a(n) 0 0 0 1 0 0 1 2 10285. b(n) 1 1 1 -1 1 1 -1 1 10286. s(n) 1 2 3 2 3 4 3 4 10287. 10288. The sequence s(n) has the remarkable property that all elements are 10289. positive and every positive integer k occurs exactly k times. 10290. 10291. Define g(t,c), with 1 ≤ c ≤ t, as the index in s(n) for which t occurs for 10292. the c'th time in s(n). 10293. E.g.: g(3,3) = 6, g(4,2) = 7 and g(54321,12345) = 1220847710. 10294. 10295. Let F(n) be the fibonacci sequence defined by: 10296. F(0)=F(1)=1 and 10297. F(n)=F(n-1)+F(n-2) for n>1. 10298. 10299. Define GF(t)=g(F(t),F(t-1)). 10300. 10301. Find ΣGF(t) for 2≤t≤45. 10302. 10303. 10304. Answer: ea0bb1fff1a51b48971762b93aeed103 10305. 10306. 10307. Problem 385 10308. =========== 10309. 10310. 10311. For any triangle T in the plane, it can be shown that there is a unique 10312. ellipse with largest area that is completely inside T. 10313. 10314. For a given n, consider triangles T such that: 10315. - the vertices of T have integer coordinates with absolute value ≤ n, and 10316. - the foci^1 of the largest-area ellipse inside T are (√13,0) and 10317. (-√13,0). 10318. Let A(n) be the sum of the areas of all such triangles. 10319. 10320. For example, if n = 8, there are two such triangles. Their vertices are 10321. (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each 10322. triangle is 36. Thus A(8) = 36 + 36 = 72. 10323. 10324. It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008. 10325. 10326. Find A(1 000 000 000). 10327. 10328. ^1The foci (plural of focus) of an ellipse are two points A and B such 10329. that for every point P on the boundary of the ellipse, AP + PB is 10330. constant. 10331. 10332. 10333. Answer: a21c033d9e119c293e51966ea78c9950 10334. 10335. 10336. Problem 386 10337. =========== 10338. 10339. 10340. Let n be an integer and S(n) be the set of factors of n. 10341. 10342. A subset A of S(n) is called an antichain of S(n) if A contains only one 10343. element or if none of the elements of A divides any of the other elements 10344. of A. 10345. 10346. For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30} 10347. {2, 5, 6} is not an antichain of S(30). 10348. {2, 3, 5} is an antichain of S(30). 10349. 10350. Let N(n) be the maximum length of an antichain of S(n). 10351. 10352. Find ΣN(n) for 1 ≤ n ≤ 10^8 10353. 10354. 10355. Answer: 528755790 10356. 10357. 10358. Problem 387 10359. =========== 10360. 10361. 10362. A Harshad or Niven number is a number that is divisible by the sum of its 10363. digits. 10364. 201 is a Harshad number because it is divisible by 3 (the sum of its 10365. digits.) 10366. When we truncate the last digit from 201, we get 20, which is a Harshad 10367. number. 10368. When we truncate the last digit from 20, we get 2, which is also a Harshad 10369. number. 10370. Let's call a Harshad number that, while recursively truncating the last 10371. digit, always results in a Harshad number a right truncatable Harshad 10372. number. 10373. 10374. Also: 10375. 201/3=67 which is prime. 10376. Let's call a Harshad number that, when divided by the sum of its digits, 10377. results in a prime a strong Harshad number. 10378. 10379. Now take the number 2011 which is prime. 10380. When we truncate the last digit from it we get 201, a strong Harshad 10381. number that is also right truncatable. 10382. Let's call such primes strong, right truncatable Harshad primes. 10383. 10384. You are given that the sum of the strong, right truncatable Harshad primes 10385. less than 10000 is 90619. 10386. 10387. Find the sum of the strong, right truncatable Harshad primes less than 10388. 10^14. 10389. 10390. 10391. Answer: a20cbd8639767decfa2c2c9955eb6be3 10392. 10393. 10394. Problem 388 10395. =========== 10396. 10397. 10398. Consider all lattice points (a,b,c) with 0 ≤ a,b,c ≤ N. 10399. 10400. From the origin O(0,0,0) all lines are drawn to the other lattice points. 10401. Let D(N) be the number of distinct such lines. 10402. 10403. You are given that D(1 000 000) = 831909254469114121. 10404. 10405. Find D(10^10). Give as your answer the first nine digits followed by the 10406. last nine digits. 10407. 10408. 10409. Answer: 2bab886c7d98d802d9249c9e12d72c25 10410. 10411. 10412. Problem 389 10413. =========== 10414. 10415. 10416. An unbiased single 4-sided die is thrown and its value, T, is noted. 10417. T unbiased 6-sided dice are thrown and their scores are added together. 10418. The sum, C, is noted. 10419. C unbiased 8-sided dice are thrown and their scores are added together. 10420. The sum, O, is noted. 10421. O unbiased 12-sided dice are thrown and their scores are added together. 10422. The sum, D, is noted. 10423. D unbiased 20-sided dice are thrown and their scores are added together. 10424. The sum, I, is noted. 10425. Find the variance of I, and give your answer rounded to 4 decimal places. 10426. 10427. 10428. Answer: 79a080d38b837547b975c97b44764dfb 10429. 10430. 10431. Problem 390 10432. =========== 10433. 10434. 10435. Consider the triangle with sides √5, √65 and √68.It can be shown that this 10436. triangle has area 9. 10437. 10438. S(n) is the sum of the areas of all triangles with sides √(1+b^2), 10439. √(1+c^2) and √(b^2+c^2) (for positive integers b and c ) that have an 10440. integral area not exceeding n. 10441. 10442. The example triangle has b=2 and c=8. 10443. 10444. S(10^6)=18018206. 10445. 10446. Find S(10^10). 10447. 10448. 10449. Answer: ed7f2fbc05a2fd2033d80de671f35ea3 10450. 10451. 10452. Problem 391 10453. =========== 10454. 10455. 10456. Let s[k] be the number of 1’s when writing the numbers from 0 to k in 10457. binary. 10458. For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101. 10459. There are seven 1’s, so s = 7. 10460. The sequence S = {s[k] : k ≥ 0} starts {0, 1, 2, 4, 5, 7, 9, 12, ...}. 10461. 10462. A game is played by two players. Before the game starts, a number n is 10463. chosen. A counter c starts at 0. At each turn, the player chooses a number 10464. from 1 to n (inclusive) and increases c by that number. The resulting 10465. value of c must be a member of S. If there are no more valid moves, the 10466. player loses. 10467. 10468. For example: 10469. Let n = 5. c starts at 0. 10470. Player 1 chooses 4, so c becomes 0 + 4 = 4. 10471. Player 2 chooses 5, so c becomes 4 + 5 = 9. 10472. Player 1 chooses 3, so c becomes 9 + 3 = 12. 10473. etc. 10474. Note that c must always belong to S, and each player can increase c by at 10475. most n. 10476. 10477. Let M(n) be the highest number the first player can choose at her first 10478. turn to force a win, and M(n) = 0 if there is no such move. For example, 10479. M(2) = 2, M(7) = 1 and M(20) = 4. 10480. 10481. Given Σ(M(n))^3 = 8150 for 1 ≤ n ≤ 20. 10482. 10483. Find Σ(M(n))^3 for 1 ≤ n ≤ 1000. 10484. 10485. 10486. Answer: b2947548d4f5c4878c5f788f9849e750 10487. 10488. 10489. Problem 392 10490. =========== 10491. 10492. 10493. A rectilinear grid is an orthogonal grid where the spacing between the 10494. gridlines does not have to be equidistant. 10495. An example of such grid is logarithmic graph paper. 10496. 10497. Consider rectilinear grids in the Cartesian coordinate system with the 10498. following properties: 10499. 10500. • The gridlines are parallel to the axes of the Cartesian coordinate 10501. system. 10502. • There are N+2 vertical and N+2 horizontal gridlines. Hence there are 10503. (N+1) x (N+1) rectangular cells. 10504. • The equations of the two outer vertical gridlines are x = -1 and x = 10505. 1. 10506. • The equations of the two outer horizontal gridlines are y = -1 and y = 10507. 1. 10508. • The grid cells are colored red if they overlap with the unit circle, 10509. black otherwise. 10510. 10511. For this problem we would like you to find the postions of the remaining N 10512. inner horizontal and N inner vertical gridlines so that the area occupied 10513. by the red cells is minimized. 10514. 10515. E.g. here is a picture of the solution for N = 10: 10516. 10517. The area occupied by the red cells for N = 10 rounded to 10 digits behind 10518. the decimal point is 3.3469640797. 10519. 10520. Find the positions for N = 400. 10521. Give as your answer the area occupied by the red cells rounded to 10 10522. digits behind the decimal point. 10523. 10524. 10525. Answer: 3268b0bc489187db3d234c097040d909 10526. 10527. 10528. Problem 393 10529. =========== 10530. 10531. 10532. An n×n grid of squares contains n^2 ants, one ant per square. 10533. All ants decide to move simultaneously to an adjacent square (usually 4 10534. possibilities, except for ants on the edge of the grid or at the corners). 10535. We define f(n) to be the number of ways this can happen without any ants 10536. ending on the same square and without any two ants crossing the same edge 10537. between two squares. 10538. 10539. You are given that f(4) = 88. 10540. Find f(10). 10541. 10542. 10543. Answer: 58e4990838fb3d1725872da30f9db748 10544. 10545. 10546. Problem 394 10547. =========== 10548. 10549. 10550. Jeff eats a pie in an unusual way. 10551. The pie is circular. He starts with slicing an initial cut in the pie 10552. along a radius. 10553. While there is at least a given fraction F of pie left, he performs the 10554. following procedure: 10555. - He makes two slices from the pie centre to any point of what is 10556. remaining of the pie border, any point on the remaining pie border equally 10557. likely. This will divide the remaining pie into three pieces. 10558. - Going counterclockwise from the initial cut, he takes the first two pie 10559. pieces and eats them. 10560. When less than a fraction F of pie remains, he does not repeat this 10561. procedure. Instead, he eats all of the remaining pie. 10562. 10563. For x ≥ 1, let E(x) be the expected number of times Jeff repeats the 10564. procedure above with F = ^1/[x]. 10565. It can be verified that E(1) = 1, E(2) ≈ 1.2676536759, and E(7.5) ≈ 10566. 2.1215732071. 10567. 10568. Find E(40) rounded to 10 decimal places behind the decimal point. 10569. 10570. 10571. Answer: f8ad575e1a03365a60b6429c3b7a64df 10572. 10573. 10574. Problem 395 10575. =========== 10576. 10577. 10578. The Pythagorean tree is a fractal generated by the following procedure: 10579. 10580. Start with a unit square. Then, calling one of the sides its base (in the 10581. animation, the bottom side is the base): 10582. 10583. 1. Attach a right triangle to the side opposite the base, with the 10584. hypotenuse coinciding with that side and with the sides in a 3-4-5 10585. ratio. Note that the smaller side of the triangle must be on the 10586. 'right' side with respect to the base (see animation). 10587. 2. Attach a square to each leg of the right triangle, with one of its 10588. sides coinciding with that leg. 10589. 3. Repeat this procedure for both squares, considering as their bases the 10590. sides touching the triangle. 10591. 10592. The resulting figure, after an infinite number of iterations, is the 10593. Pythagorean tree. 10594. 10595. It can be shown that there exists at least one rectangle, whose sides are 10596. parallel to the largest square of the Pythagorean tree, which encloses the 10597. Pythagorean tree completely. 10598. 10599. Find the smallest area possible for such a bounding rectangle, and give 10600. your answer rounded to 10 decimal places. 10601. 10602. 10603. p_395_pythagorean.gif 10604. Answer: 505048b0c619161d05b9b3e492f3edc3 10605. 10606. 10607. Problem 396 10608. =========== 10609. 10610. 10611. For any positive integer n, the nth weak Goodstein sequence {g, g, 10612. g, ...} is defined as: 10613. 10614. • g = n 10615. • for k > 1, g[k] is obtained by writing g[k-1] in base k, interpreting 10616. it as a base k + 1 number, and subtracting 1. 10617. 10618. The sequence terminates when g[k] becomes 0. 10619. 10620. For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}: 10621. 10622. • g = 6. 10623. • g = 11 since 6 = 110, 110 = 12, and 12 - 1 = 11. 10624. • g = 17 since 11 = 102, 102 = 18, and 18 - 1 = 17. 10625. • g = 25 since 17 = 101, 101 = 26, and 26 - 1 = 25. 10626. 10627. and so on. 10628. 10629. It can be shown that every weak Goodstein sequence terminates. 10630. 10631. Let G(n) be the number of nonzero elements in the nth weak Goodstein 10632. sequence. 10633. It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381. 10634. It can also be verified that ΣG(n) = 2517 for 1 ≤ n < 8. 10635. 10636. Find the last 9 digits of ΣG(n) for 1 ≤ n < 16. 10637. 10638. 10639. Answer: 173214653 10640. 10641. 10642. Problem 397 10643. =========== 10644. 10645. 10646. On the parabola y = x^2/k, three points A(a, a^2/k), B(b, b^2/k) and C(c, 10647. c^2/k) are chosen. 10648. 10649. Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such 10650. that at least one angle of the triangle ABC is 45-degree, with 1 ≤ k ≤ K 10651. and -X ≤ a < b < c ≤ X. 10652. 10653. For example, F(1, 10) = 41 and F(10, 100) = 12492. 10654. Find F(10^6, 10^9). 10655. 10656. 10657. Answer: 07f769df9543bc05e6318878c34d074d 10658. 10659. 10660. Problem 398 10661. =========== 10662. 10663. 10664. Inside a rope of length n, n-1 points are placed with distance 1 from each 10665. other and from the endpoints. Among these points, we choose m-1 points at 10666. random and cut the rope at these points to create m segments. 10667. 10668. Let E(n, m) be the expected length of the second-shortest segment.For 10669. example, E(3, 2) = 2 and E(8, 3) = 16/7.Note that if multiple segments 10670. have the same shortest length the length of the second-shortest segment is 10671. defined as the same as the shortest length. 10672. 10673. Find E(10^7, 100).Give your answer rounded to 5 decimal places behind the 10674. decimal point. 10675. 10676. 10677. Answer: fa0a25d62fa225e05fd8736713a9bfc0 10678. 10679. 10680. Problem 399 10681. =========== 10682. 10683. 10684. The first 15 fibonacci numbers are: 10685. 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610. 10686. It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and 10687. 144 is divisible by 4 and by 9. 10688. So the first 13 squarefree fibonacci numbers are: 10689. 1,1,2,3,5,13,21,34,55,89,233,377 and 610. 10690. 10691. The 200th squarefree fibonacci number 10692. is:971183874599339129547649988289594072811608739584170445. 10693. The last sixteen digits of this number are: 1608739584170445 and in 10694. scientific notation this number can be written as 9.7e53. 10695. 10696. Find the 100 000 000th squarefree fibonacci number. 10697. Give as your answer its last sixteen digits followed by a comma followed 10698. by the number in scientific notation (rounded to one digit after the 10699. decimal point). 10700. For the 200th squarefree number the answer would have been: 10701. 1608739584170445,9.7e53 10702. 10703. Note: 10704. For this problem, assume that for every prime p, the first fibonacci 10705. number divisible by p is not divisible by p^2 (this is part of Wall's 10706. conjecture). This has been verified for primes ≤ 3·10^15, but has not been 10707. proven in general. 10708. If it happens that the conjecture is false, then the accepted answer to 10709. this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci 10710. number, rather it represents only a lower bound for that number. 10711. 10712. 10713. Answer: a0819cfe3be6a04645b8d4fe2345e184 10714. 10715. 10716. Problem 400 10717. =========== 10718. 10719. 10720. A Fibonacci tree is a binary tree recursively defined as: 10721. 10722. • T(0) is the empty tree. 10723. • T(1) is the binary tree with only one node. 10724. • T(k) consists of a root node that has T(k-1) and T(k-2) as children. 10725. 10726. On such a tree two players play a take-away game. On each turn a player 10727. selects a node and removes that node along with the subtree rooted at that 10728. node. 10729. The player who is forced to take the root node of the entire tree loses. 10730. 10731. Here are the winning moves of the first player on the first turn for T(k) 10732. from k=1 to k=6. 10733. 10734. Let f(k) be the number of winning moves of the first player (i.e. the 10735. moves for which the second player has no winning strategy) on the first 10736. turn of the game when this game is played on T(k). 10737. 10738. For example, f(5) = 1 and f(10) = 17. 10739. 10740. Find f(10000). Give the last 18 digits of your answer. 10741. 10742. 10743. Answer: 60aa790c07af1446c1e2deba72543a1f 10744. 10745. 10746. Problem 401 10747. =========== 10748. 10749. 10750. The divisors of 6 are 1,2,3 and 6. 10751. The sum of the squares of these numbers is 1+4+9+36=50. 10752. 10753. Let sigma2(n) represent the sum of the squares of the divisors of n.Thus 10754. sigma2(6)=50. 10755. 10756. Let SIGMA2 represent the summatory function of sigma2, that is 10757. SIGMA2(n)=∑sigma2(i) for i=1 to n. 10758. The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113. 10759. 10760. Find SIGMA2(10^15) modulo 10^9. 10761. 10762. 10763. Answer: 982a249d8b45ef10c98c32dabac00751 10764. 10765. 10766. Problem 402 10767. =========== 10768. 10769. 10770. It can be shown that the polynomial n^4 + 4n^3 + 2n^2 + 5n is a multiple 10771. of 6 for every integer n. It can also be shown that 6 is the largest 10772. integer satisfying this property. 10773. 10774. Define M(a, b, c) as the maximum m such that n^4 + an^3 + bn^2 + cn is a 10775. multiple of m for all integers n. For example, M(4, 2, 5) = 6. 10776. 10777. Also, define S(N) as the sum of M(a, b, c) for all 0 < a, b, c ≤ N. 10778. 10779. We can verify that S(10) = 1972 and S(10000) = 2024258331114. 10780. 10781. Let F[k] be the Fibonacci sequence: 10782. F = 0, F = 1 and 10783. F[k] = F[k-1] + F[k-2] for k ≥ 2. 10784. 10785. Find the last 9 digits of Σ S(F[k]) for 2 ≤ k ≤ 1234567890123. 10786. 10787. 10788. Answer: fa7ae8e9243f01b0eac10ec5aaff1f42 10789. 10790. 10791. Problem 403 10792. =========== 10793. 10794. 10795. For integers a and b, we define D(a, b) as the domain enclosed by the 10796. parabola y = x^2 and the line y = a·x + b: 10797. D(a, b) = { (x, y) | x^2 ≤ y ≤ a·x + b }. 10798. 10799. L(a, b) is defined as the number of lattice points contained in D(a, b). 10800. For example, L(1, 2) = 8 and L(2, -1) = 1. 10801. 10802. We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such 10803. that the area of D(a, b) is a rational number and |a|,|b| ≤ N. 10804. We can verify that S(5) = 344 and S(100) = 26709528. 10805. 10806. Find S(10^12). Give your answer mod 10^8. 10807. 10808. 10809. Answer: 18224771 10810. 10811. 10812. Problem 404 10813. =========== 10814. 10815. 10816. E[a] is an ellipse with an equation of the form x^2 + 4y^2 = 4a^2. 10817. E[a]' is the rotated image of E[a] by θ degrees counterclockwise around 10818. the origin O(0, 0) for 0° < θ < 90°. 10819. 10820. b is the distance to the origin of the two intersection points closest to 10821. the origin and c is the distance of the two other intersection points. 10822. We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a, 10823. b and c are positive integers. 10824. For example, (209, 247, 286) is a canonical ellipsoidal triplet. 10825. 10826. Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b, 10827. c) for a ≤ N. 10828. It can be verified that C(10^3) = 7, C(10^4) = 106 and C(10^6) = 11845. 10829. 10830. Find C(10^17). 10831. 10832. 10833. p_404_c_ellipse.gif 10834. Answer: 2d1bc4b93bbc19d9e70c5b04338dea2e 10835. 10836. 10837. Problem 405 10838. =========== 10839. 10840. 10841. We wish to tile a rectangle whose length is twice its width. 10842. Let T(0) be the tiling consisting of a single rectangle. 10843. For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the 10844. following manner: 10845. 10846. The following animation demonstrates the tilings T(n) for n from 0 to 5: 10847. 10848. Let f(n) be the number of points where four tiles meet in T(n). 10849. For example, f(1) = 0, f(4) = 82 and f(10^9) mod 17^7 = 126897180. 10850. 10851. Find f(10^k) for k = 10^18, give your answer modulo 17^7. 10852. 10853. 10854. p_405_tile1.png 10855. p_405_tile2.gif 10856. Answer: 93b712426b768586f88d0bfe597842e6 10857. 10858. 10859. Problem 406 10860. =========== 10861. 10862. 10863. We are trying to find a hidden number selected from the set of integers 10864. {1, 2, ..., n} by asking questions. Each number (question) we ask, we get 10865. one of three possible answers: 10866. 10867. • "Your guess is lower than the hidden number" (and you incur a cost of 10868. a), or 10869. • "Your guess is higher than the hidden number" (and you incur a cost of 10870. b), or 10871. • "Yes, that's it!" (and the game ends). 10872. 10873. Given the value of n, a, and b, an optimal strategy minimizes the total 10874. cost for the worst possible case. 10875. 10876. For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2" 10877. as our first question. 10878. 10879. If we are told that 2 is higher than the hidden number (for a cost of 10880. b=3), then we are sure that "1" is the hidden number (for a total cost of 10881. 3). 10882. If we are told that 2 is lower than the hidden number (for a cost of a=2), 10883. then our next question will be "4". 10884. If we are told that 4 is higher than the hidden number (for a cost of 10885. b=3), then we are sure that "3" is the hidden number (for a total cost of 10886. 2+3=5). 10887. If we are told that 4 is lower than the hidden number (for a cost of a=2), 10888. then we are sure that "5" is the hidden number (for a total cost of 10889. 2+2=4). 10890. Thus, the worst-case cost achieved by this strategy is 5. It can also be 10891. shown that this is the lowest worst-case cost that can be achieved. So, in 10892. fact, we have just described an optimal strategy for the given values of 10893. n, a, and b. 10894. 10895. Let C(n, a, b) be the worst-case cost achieved by an optimal strategy for 10896. the given values of n, a, and b. 10897. 10898. Here are a few examples: 10899. C(5, 2, 3) = 5 10900. C(500, √2, √3) = 13.22073197... 10901. C(20000, 5, 7) = 82 10902. C(2000000, √5, √7) = 49.63755955... 10903. 10904. Let F[k] be the Fibonacci numbers: F[k] = F[k-1] + F[k-2] with base cases 10905. F = F = 1. 10906. Find ∑[1≤k≤30] C(10^12, √k, √F[k]), and give your answer rounded to 8 10907. decimal places behind the decimal point. 10908. 10909. 10910. Answer: 0766b1ee975f5674d30fd6c3c934c6e0 10911. 10912. 10913. Problem 407 10914. =========== 10915. 10916. 10917. If we calculate a^2 mod 6 for 0 ≤ a ≤ 5 we get: 0,1,4,3,4,1. 10918. 10919. The largest value of a such that a^2 ≡ a mod 6 is 4. 10920. Let's call M(n) the largest value of a < n such that a^2 ≡ a (mod n). 10921. So M(6) = 4. 10922. 10923. Find ∑M(n) for 1 ≤ n ≤ 10^7. 10924. 10925. 10926. Answer: f4da34a4b357123cb142739a52e010f2 10927. 10928. 10929. Problem 408 10930. =========== 10931. 10932. 10933. Let's call a lattice point (x, y) inadmissible if x, y and x + y are all 10934. positive perfect squares. 10935. For example, (9, 16) is inadmissible, while (0, 4), (3, 1) and (9, 4) are 10936. not. 10937. 10938. Consider a path from point (x, y) to point (x, y) using only 10939. unit steps north or east. 10940. Let's call such a path admissible if none of its intermediate points are 10941. inadmissible. 10942. 10943. Let P(n) be the number of admissible paths from (0, 0) to (n, n). 10944. It can be verified that P(5) = 252, P(16) = 596994440 and P(1000) mod 10945. 1 000 000 007 = 341920854. 10946. 10947. Find P(10 000 000) mod 1 000 000 007. 10948. 10949. 10950. Answer: 2c09e247c6144c16cae2358d316affd9 10951. 10952. 10953. Problem 409 10954. =========== 10955. 10956. 10957. Let n be a positive integer. Consider nim positions where: 10958. 10959. • There are n non-empty piles. 10960. • Each pile has size less than 2^n. 10961. • No two piles have the same size. 10962. 10963. Let W(n) be the number of winning nim positions satisfying the 10964. aboveconditions (a position is winning if the first player has a winning 10965. strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360 10966. and W(100) mod 1 000 000 007 = 384777056. 10967. 10968. Find W(10 000 000) mod 1 000 000 007. 10969. 10970. 10971. Answer: 56c32e75a2656ec08ce177089bda2f53 10972. 10973. 10974. Problem 410 10975. =========== 10976. 10977. 10978. Let C be the circle with radius r, x^2 + y^2 = r^2. We choose two points 10979. P(a, b) and Q(-a, c) so that the line passing through P and Q is tangent 10980. to C. 10981. 10982. For example, the quadruplet (r, a, b, c) = (2, 6, 2, -7) satisfies this 10983. property. 10984. 10985. Let F(R, X) be the number of the integer quadruplets (r, a, b, c) with 10986. this property, and with 0 < r ≤ R and 0 < a ≤ X. 10987. 10988. We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384. 10989. Find F(10^8, 10^9) + F(10^9, 10^8). 10990. 10991. 10992. Answer: 45826f3a23aa321f97acb1d2a8f2170b 10993. 10994. 10995. Problem 411 10996. =========== 10997. 10998. 10999. Let n be a positive integer. Suppose there are stations at the coordinates 11000. (x, y) = (2^i mod n, 3^i mod n) for 0 ≤ i ≤ 2n. We will consider stations 11001. with the same coordinates as the same station. 11002. 11003. We wish to form a path from (0, 0) to (n, n) such that the x and y 11004. coordinates never decrease. 11005. Let S(n) be the maximum number of stations such a path can pass through. 11006. 11007. For example, if n = 22, there are 11 distinct stations, and a valid path 11008. can pass through at most 5 stations. Therefore, S(22) = 5.The case is 11009. illustrated below, with an example of an optimal path: 11010. 11011. It can also be verified that S(123) = 14 and S(10000) = 48. 11012. 11013. Find ∑ S(k^5) for 1 ≤ k ≤ 30. 11014. 11015. 11016. Answer: 9936352 11017. 11018. 11019. Problem 412 11020. =========== 11021. 11022. 11023. For integers m, n (0 ≤ n < m), let L(m, n) be an m×m grid with the 11024. top-right n×n grid removed. 11025. 11026. For example, L(5, 3) looks like this: 11027. 11028. We want to number each cell of L(m, n) with consecutive integers 1, 2, 3, 11029. ... such that the number in every cell is smaller than the number below it 11030. and to the left of it. 11031. 11032. For example, here are two valid numberings of L(5, 3): 11033. 11034. Let LC(m, n) be the number of valid numberings of L(m, n). 11035. It can be verified that LC(3, 0) = 42, LC(5, 3) = 250250, LC(6, 3) = 11036. 406029023400 and LC(10, 5) mod 76543217 = 61251715. 11037. 11038. Find LC(10000, 5000) mod 76543217. 11039. 11040. 11041. Answer: 38788800 11042. 11043. 11044. Problem 413 11045. =========== 11046. 11047. 11048. We say that a d-digit positive number (no leading zeros) is a one-child 11049. number if exactly one of its sub-strings is divisible by d. 11050. 11051. For example, 5671 is a 4-digit one-child number. Among all its sub-strings 11052. 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4. 11053. Similarly, 104 is a 3-digit one-child number because only 0 is divisible 11054. by 3. 11055. 1132451 is a 7-digit one-child number because only 245 is divisible by 7. 11056. 11057. Let F(N) be the number of the one-child numbers less than N. 11058. We can verify that F(10) = 9, F(10^3) = 389 and F(10^7) = 277674. 11059. 11060. Find F(10^19). 11061. 11062. 11063. Answer: 569ad33af889215704df5a9e278aa004 11064. 11065. 11066. Problem 414 11067. =========== 11068. 11069. 11070. 6174 is a remarkable number; if we sort its digits in increasing order and 11071. subtract that number from the number you get when you sort the digits in 11072. decreasing order, we get 7641-1467=6174. 11073. Even more remarkable is that if we start from any 4 digit number and 11074. repeat this process of sorting and subtracting, we'll eventually end up 11075. with 6174 or immediately with 0 if all digits are equal. 11076. This also works with numbers that have less than 4 digits if we pad the 11077. number with leading zeroes until we have 4 digits. 11078. E.g. let's start with the number 0837: 11079. 8730-0378=8352 11080. 8532-2358=6174 11081. 11082. 6174 is called the Kaprekar constant. The process of sorting and 11083. subtracting and repeating this until either 0 or the Kaprekar constant is 11084. reached is called the Kaprekar routine. 11085. 11086. We can consider the Kaprekar routine for other bases and number of digits. 11087. Unfortunately, it is not guaranteed a Kaprekar constant exists in all 11088. cases; either the routine can end up in a cycle for some input numbers or 11089. the constant the routine arrives at can be different for different input 11090. numbers. 11091. However, it can be shown that for 5 digits and a base b = 6t+3≠9, a 11092. Kaprekar constant exists. 11093. E.g. base 15: (10,4,14,9,5) 11094. base 21: (14,6,20,13,7) 11095. 11096. Define C[b] to be the Kaprekar constant in base b for 5 digits.Define the 11097. function sb(i) to be 11098. 11099. • 0 if i = C[b] or if i written in base b consists of 5 identical digits 11100. • the number of iterations it takes the Kaprekar routine in base b to 11101. arrive at C[b], otherwise 11102. 11103. Note that we can define sb(i) for all integers i < b^5. If i written in 11104. base b takes less than 5 digits, the number is padded with leading zero 11105. digits until we have 5 digits before applying the Kaprekar routine. 11106. 11107. Define S(b) as the sum of sb(i) for 0 < i < b^5. 11108. E.g. S(15) = 5274369 11109. S(111) = 400668930299 11110. 11111. Find the sum of S(6k+3) for 2 ≤ k ≤ 300. 11112. Give the last 18 digits as your answer. 11113. 11114. 11115. Answer: 42f095bdfd71e1ae4ae0ceead1bb1802 11116. 11117. 11118. Problem 415 11119. =========== 11120. 11121. 11122. A set of lattice points S is called a titanic set if there exists a line 11123. passing through exactly two points in S. 11124. 11125. An example of a titanic set is S = {(0, 0), (0, 1), (0, 2), (1, 1), (2, 11126. 0), (1, 0)}, where the line passing through (0, 1) and (2, 0) does not 11127. pass through any other point in S. 11128. 11129. On the other hand, the set {(0, 0), (1, 1), (2, 2), (4, 4)} is not a 11130. titanic set since the line passing through any two points in the set also 11131. passes through the other two. 11132. 11133. For any positive integer N, let T(N) be the number of titanic sets S whose 11134. every point (x, y) satisfies 0 ≤ x, y ≤ N.It can be verified that T(1) = 11135. 11, T(2) = 494, T(4) = 33554178, T(111) mod 10^8 = 13500401 and 11136. T(10^5) mod 10^8 = 63259062. 11137. 11138. Find T(10^11) mod 10^8. 11139. 11140. 11141. Answer: 55859742 11142. 11143. 11144. Problem 416 11145. =========== 11146. 11147. 11148. A row of n squares contains a frog in the leftmost square. By successive 11149. jumps the frog goes to the rightmost square and then back to the leftmost 11150. square. On the outward trip he jumps one, two or three squares to the 11151. right, and on the homeward trip he jumps to the left in a similar manner. 11152. He cannot jump outside the squares. He repeats the round-trip travel m 11153. times. 11154. 11155. Let F(m, n) be the number of the ways the frog can travel so that at most 11156. one square remains unvisited. 11157. For example, F(1, 3) = 4, F(1, 4) = 15, F(1, 5) = 46, F(2, 3) = 16 and 11158. F(2, 100) mod 10^9 = 429619151. 11159. 11160. Find the last 9 digits of F(10, 10^12). 11161. 11162. 11163. Answer: 6f398386fdfec57ac166d4970c2bcad2 11164. 11165. 11166. Problem 417 11167. =========== 11168. 11169. 11170. A unit fraction contains 1 in the numerator. The decimal representation of 11171. the unit fractions with denominators 2 to 10 are given: 11172. 11173. 1/2 = 0.5 11174. 1/3 = 0.(3) 11175. 1/4 = 0.25 11176. 1/5 = 0.2 11177. 1/6 = 0.1(6) 11178. 1/7 = 0.(142857) 11179. 1/8 = 0.125 11180. 1/9 = 0.(1) 11181. 1/10 = 0.1 11182. 11183. Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can 11184. be seen that 1/7 has a 6-digit recurring cycle. 11185. 11186. Unit fractions whose denominator has no other prime factors than 2 and/or 11187. 5 are not considered to have a recurring cycle. 11188. We define the length of the recurring cycle of those unit fractions as 0. 11189. 11190. Let L(n) denote the length of the recurring cycle of 1/n.You are given 11191. that ∑L(n) for 3 ≤ n ≤ 1 000 000 equals 55535191115. 11192. 11193. Find ∑L(n) for 3 ≤ n ≤ 100 000 000 11194. 11195. 11196. Answer: 93a7df08c972f1e7788516d056a7e016 11197. 11198. 11199. Problem 418 11200. =========== 11201. 11202. 11203. Let n be a positive integer. An integer triple (a, b, c) is called a 11204. factorisation triple of n if: 11205. 11206. • 1 ≤ a ≤ b ≤ c 11207. • a·b·c = n. 11208. 11209. Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n 11210. which minimises c / a. One can show that this triple is unique. 11211. 11212. For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872. 11213. 11214. Find f(43!). 11215. 11216. 11217. Answer: b032468ddb4847d8a2273789379753f5 11218. 11219. 11220. Problem 419 11221. =========== 11222. 11223. 11224. The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, 11225. 1113213211, ... 11226. The sequence starts with 1 and all other members are obtained by 11227. describing the previous member in terms of consecutive digits. 11228. It helps to do this out loud: 11229. 1 is 'one one' → 11 11230. 11 is 'two ones' → 21 11231. 21 is 'one two and one one' → 1211 11232. 1211 is 'one one, one two and two ones' → 111221 11233. 111221 is 'three ones, two twos and one one' → 312211 11234. ... 11235. 11236. Define A(n), B(n) and C(n) as the number of ones, twos and threes in the 11237. n'th element of the sequence respectively. 11238. One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625. 11239. 11240. Find A(n), B(n) and C(n) for n = 10^12. 11241. Give your answer modulo 2^30 and separate your values for A, B and C by a 11242. comma. 11243. E.g. for n = 40 the answer would be 31254,20259,11625 11244. 11245. 11246. Answer: b27db655498b3d64ad4338fcdc9d178f 11247. 11248. 11249. Problem 420 11250. =========== 11251. 11252. 11253. A positive integer matrix is a matrix whose elements are all positive 11254. integers. 11255. Some positive integer matrices can be expressed as a square of a positive 11256. integer matrix in two different ways. Here is an example: 11257. 11258. We define F(N) as the number of the 2x2 positive integer matrices which 11259. have a trace less than N and which can be expressed as a square of a 11260. positive integer matrix in two different ways. 11261. We can verify that F(50) = 7 and F(1000) = 1019. 11262. 11263. Find F(10^7). 11264. 11265. 11266. p_420_matrix.gif 11267. Answer: 145159332 11268. 11269. 11270. Problem 421 11271. =========== 11272. 11273. 11274. Numbers of the form n^15+1 are composite for every integer n > 1. 11275. For positive integers n and m let s(n,m) be defined as the sum of the 11276. distinct prime factors of n^15+1 not exceeding m. 11277. 11278. E.g. 2^15+1 = 3×3×11×331. 11279. So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345. 11280. 11281. Also 10^15+1 = 7×11×13×211×241×2161×9091. 11282. So s(10,100) = 31 and s(10,1000) = 483. 11283. 11284. Find &Sum; s(n,10^8) for 1 ≤ n ≤ 10^11. 11285. 11286. 11287. Answer: 481fcc5ff16ccf1645fb136c123ed660 11288. 11289. 11290. Problem 422 11291. =========== 11292. 11293. 11294. Let H be the hyperbola defined by the equation 12x^2 + 7xy - 12y^2 = 625. 11295. 11296. Next, define X as the point (7, 1). It can be seen that X is in H. 11297. 11298. Now we define a sequence of points in H, {P[i] : i ≥ 1}, as: 11299. 11300. • P = (13, 61/4). 11301. • P = (-43/6, -4). 11302. • For i > 2, P[i] is the unique point in H that is different from P[i-1] 11303. and such that line P[i]P[i-1] is parallel to line P[i-2]X. It can be 11304. shown that P[i] is well-defined, and that its coordinates are always 11305. rational. 11306. 11307. You are given that P = (-19/2, -229/24), P = (1267/144, -37/12) and 11308. P = (17194218091/143327232, 274748766781/1719926784). 11309. 11310. Find P[n] for n = 11^14 in the following format: 11311. If P[n] = (a/b, c/d) where the fractions are in lowest terms and the 11312. denominators are positive, then the answer is (a + b + c + d) mod 11313. 1 000 000 007. 11314. 11315. For n = 7, the answer would have been: 806236837. 11316. 11317. 11318. Answer: 92060460 11319. 11320. 11321. Problem 423 11322. =========== 11323. 11324. 11325. Let n be a positive integer. 11326. A 6-sided die is thrown n times. Let c be the number of pairs of 11327. consecutive throws that give the same value. 11328. 11329. For example, if n = 7 and the values of the die throws are 11330. (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the 11331. same value: 11332. (1,1,5,6,6,6,3) 11333. (1,1,5,6,6,6,3) 11334. (1,1,5,6,6,6,3) 11335. Therefore, c = 3 for (1,1,5,6,6,6,3). 11336. 11337. Define C(n) as the number of outcomes of throwing a 6-sided die n times 11338. such that c does not exceed π(n).^1 11339. For example, C(3) = 216, C(4) = 1290, C(11) = 361912500 and C(24) = 11340. 4727547363281250000. 11341. 11342. Define S(L) as ∑ C(n) for 1 ≤ n ≤ L. 11343. For example, S(50) mod 1 000 000 007 = 832833871. 11344. 11345. Find S(50 000 000) mod 1 000 000 007. 11346. 11347. ^1 π denotes the prime-counting function, i.e. π(n) is the number of 11348. primes ≤ n. 11349. 11350. 11351. Answer: e2add9d46ebd8ba59a07dca791cd629b 11352. 11353. 11354. Problem 424 11355. =========== 11356. 11357. 11358. The above is an example of a cryptic kakuro (also known as cross sums, or 11359. even sums cross) puzzle, with its final solution on the right. (The common 11360. rules of kakuro puzzles can be found easily on numerous internet sites. 11361. Other related information can also be currently found at krazydad.com 11362. whose author has provided the puzzle data for this challenge.) 11363. 11364. The downloadable text file (kakuro200.txt) contains the description of 11365. 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file 11366. is the above example which is coded as follows: 11367. 11368. 6,X,X,(vCC),(vI),X,X,X,(hH),B,O,(vCA),(vJE),X,(hFE,vD),O,O,O,O,(hA),O,I,(hJC,vB),O,O,(hJC),H,O,O,O,X,X,X,(hJE),O,O,X 11369. 11370. The first character is a numerical digit indicating the size of the 11371. information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7 11372. (for a 6x6 puzzle) followed by a comma (,). The extra top line and left 11373. column are needed to insert information. 11374. 11375. The content of each cell is then described and followed by a comma, going 11376. left to right and starting with the top line. 11377. X = Gray cell, not required to be filled by a digit. 11378. O (upper case letter)= White empty cell to be filled by a digit. 11379. A = Or any one of the upper case letters from A to J to be replaced by its 11380. equivalent digit in the solved puzzle. 11381. ( ) = Location of the encrypted sums. Horizontal sums are preceded by a 11382. lower case "h" and vertical sums are preceded by a lower case "v". Those 11383. are followed by one or two upper case letters depending if the sum is a 11384. single digit or double digit one. For double digit sums, the first letter 11385. would be for the "tens" and the second one for the "units". When the cell 11386. must contain information for both a horizontal and a vertical sum, the 11387. first one is always for the horizontal sum and the two are separated by a 11388. comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets 11389. is also immediately followed by a comma. 11390. 11391. The description of the last cell is followed by a Carriage Return/Line 11392. Feed (CRLF) instead of a comma. 11393. 11394. The required answer to each puzzle is based on the value of each letter 11395. necessary to arrive at the solution and according to the alphabetical 11396. order. As indicated under the example puzzle, its answer would be 11397. 8426039571. At least 9 out of the 10 encrypting letters are always part of 11398. the problem description. When only 9 are given, the missing one must be 11399. assigned the remaining digit. 11400. 11401. You are given that the sum of the answers for the first 10 puzzles in the 11402. file is 64414157580. 11403. 11404. Find the sum of the answers for the 200 puzzles. 11405. 11406. 11407. Visible links 11408. 1. http://krazydad.com/ 11409. 2. kakuro200.txt 11410. Answer: 1059760019628 11411. 11412. 11413. Problem 425 11414. =========== 11415. 11416. 11417. Two positive numbers A and B are said to be connected (denoted by "A ↔ B") 11418. if one of these conditions holds: 11419. (1) A and B have the same length and differ in exactly one digit; for 11420. example, 123 ↔ 173. 11421. (2) Adding one digit to the left of A (or B) makes B (or A); for example, 11422. 23 ↔ 223 and 123 ↔ 23. 11423. 11424. We call a prime P a 2's relative if there exists a chain of connected 11425. primes between 2 and P and no prime in the chain exceeds P. 11426. 11427. For example, 127 is a 2's relative. One of the possible chains is shown 11428. below: 11429. 2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127 11430. However, 11 and 103 are not 2's relatives. 11431. 11432. Let F(N) be the sum of the primes ≤ N which are not 2's relatives. 11433. We can verify that F(10^3) = 431 and F(10^4) = 78728. 11434. 11435. Find F(10^7). 11436. 11437. 11438. Answer: 3d229894ba4c585138125e802af2d06e 11439. 11440. 11441. Problem 426 11442. =========== 11443. 11444. 11445. Consider an infinite row of boxes. Some of the boxes contain a ball. For 11446. example, an initial configuration of 2 consecutive occupied boxes followed 11447. by 2 empty boxes, 2 occupied boxes, 1 empty box, and 2 occupied boxes can 11448. be denoted by the sequence (2, 2, 2, 1, 2), in which the number of 11449. consecutive occupied and empty boxes appear alternately. 11450. 11451. A turn consists of moving each ball exactly once according to the 11452. following rule: Transfer the leftmost ball which has not been moved to the 11453. nearest empty box to its right. 11454. 11455. After one turn the sequence (2, 2, 2, 1, 2) becomes (2, 2, 1, 2, 3) as can 11456. be seen below; note that we begin the new sequence starting at the first 11457. occupied box. 11458. 11459. A system like this is called a Box-Ball System or BBS for short. 11460. 11461. It can be shown that after a sufficient number of turns, the system 11462. evolves to a state where the consecutive numbers of occupied boxes is 11463. invariant. In the example below, the consecutive numbers of occupied boxes 11464. evolves to [1, 2, 3]; we shall call this the final state. 11465. 11466. We define the sequence {t[i]}: 11467. 11468. • s = 290797 11469. • s[k+1] = s[k]^2 mod 50515093 11470. • t[k] = (s[k] mod 64) + 1 11471. 11472. Starting from the initial configuration (t, t, …, t), the final 11473. state becomes [1, 3, 10, 24, 51, 75]. 11474. Starting from the initial configuration (t, t, …, t[10 000 000]), 11475. find the final state. 11476. Give as your answer the sum of the squares of the elements of the final 11477. state. For example, if the final state is [1, 2, 3] then 14 ( = 1^2 + 2^2 11478. + 3^2) is your answer. 11479. 11480. 11481. p_426_baxball1.gif 11482. p_426_baxball2.gif 11483. Answer: b5d8157a351482da47da0512ca374007 11484. 11485. 11486. Problem 427 11487. =========== 11488. 11489. 11490. A sequence of integers S = {s[i]} is called an n-sequence if it has n 11491. elements and each element s[i] satisfies 1 ≤ s[i] ≤ n. Thus there are n^n 11492. distinct n-sequences in total.For example, the sequence S = {1, 5, 5, 10, 11493. 7, 7, 7, 2, 3, 7} is a 10-sequence. 11494. 11495. For any sequence S, let L(S) be the length of the longest contiguous 11496. subsequence of S with the same value.For example, for the given sequence S 11497. above, L(S) = 3, because of the three consecutive 7's. 11498. 11499. Let f(n) = ∑ L(S) for all n-sequences S. 11500. 11501. For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121. 11502. 11503. Find f(7 500 000) mod 1 000 000 009. 11504. 11505. 11506. Answer: 97138867 11507. 11508. 11509. Problem 428 11510. =========== 11511. 11512. 11513. Let a, b and c be positive numbers. 11514. Let W, X, Y, Z be four collinear points where |WX| = a, |XY| = b, |YZ| = c 11515. and |WZ| = a + b + c. 11516. Let C[in] be the circle having the diameter XY. 11517. Let C[out] be the circle having the diameter WZ. 11518. 11519. The triplet (a, b, c) is called a necklace triplet if you can place k ≥ 3 11520. distinct circles C, C, ..., C[k] such that: 11521. 11522. • C[i] has no common interior points with any C[j] for 1 ≤ i, j ≤ k and 11523. i ≠ j, 11524. • C[i] is tangent to both C[in] and C[out] for 1 ≤ i ≤ k, 11525. • C[i] is tangent to C[i+1] for 1 ≤ i < k, and 11526. • C[k] is tangent to C. 11527. 11528. For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can 11529. be shown that (2, 2, 5) is not. 11530. 11531. Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c 11532. are positive integers, and b ≤ n.For example, T(1) = 9, T(20) = 732 and 11533. T(3000) = 438106. 11534. 11535. Find T(1 000 000 000). 11536. 11537. 11538. Answer: c6010c109b66b34bf3594e63eb58b446 11539. 11540. 11541. Problem 429 11542. =========== 11543. 11544. 11545. A unitary divisor d of a number n is a divisor of n that has the property 11546. gcd(d, n/d) = 1. 11547. The unitary divisors of 4! = 24 are 1, 3, 8 and 24. 11548. The sum of their squares is 1^2 + 3^2 + 8^2 + 24^2 = 650. 11549. 11550. Let S(n) represent the sum of the squares of the unitary divisors of n. 11551. Thus S(4!)=650. 11552. 11553. Find S(100 000 000!) modulo 1 000 000 009. 11554. 11555. 11556. Answer: 98792821 11557. 11558. 11559. Problem 430 11560. =========== 11561. 11562. 11563. N disks are placed in a row, indexed 1 to N from left to right. 11564. Each disk has a black side and white side. Initially all disks show their 11565. white side. 11566. 11567. At each turn, two, not necessarily distinct, integers A and B between 1 11568. and N (inclusive) are chosen uniformly at random. 11569. All disks with an index from A to B (inclusive) are flipped. 11570. 11571. The following example shows the case N = 8. At the first turn A = 5 and B 11572. = 2, and at the second turn A = 4 and B = 6. 11573. 11574. Let E(N, M) be the expected number of disks that show their white side 11575. after M turns. 11576. We can verify that E(3, 1) = 10/9, E(3, 2) = 5/3, E(10, 4) ≈ 5.157 and 11577. E(100, 10) ≈ 51.893. 11578. 11579. Find E(10^10, 4000). 11580. Give your answer rounded to 2 decimal places behind the decimal point. 11581. 11582. 11583. p_430_flips.gif 11584. Answer: 32b0825d7a110a1a220e80629c413411 11585. 11586. 11587. Problem 431 11588. =========== 11589. 11590. 11591. Fred the farmer arranges to have a new storage silo installed on his farm 11592. and having an obsession for all things square he is absolutely devastated 11593. when he discovers that it is circular. Quentin, the representative from 11594. the company that installed the silo, explains that they only manufacture 11595. cylindrical silos, but he points out that it is resting on a square base. 11596. Fred is not amused and insists that it is removed from his property. 11597. 11598. Quick thinking Quentin explains that when granular materials are delivered 11599. from above a conical slope is formed and the natural angle made with the 11600. horizontal is called the angle of repose. For example if the angle of 11601. repose,\alpha = 30$degrees, and grain is delivered at the centre of the 11602. silo then a perfect cone will form towards the top of the cylinder. In the 11603. case of this silo, which has a diameter of 6m, the amount of space wasted 11604. would be approximately 32.648388556 m^3. However, if grain is delivered at 11605. a point on the top which has a horizontal distance of$x$metres from the 11606. centre then a cone with a strangely curved and sloping base is formed. He 11607. shows Fred a picture. 11608. 11609. We shall let the amount of space wasted in cubic metres be given by 11610.$V(x)$. If$x = 1.114785284$, which happens to have three squared decimal 11611. places, then the amount of space wasted,$V(1.114785284) \approx 36$. 11612. Given the range of possible solutions to this problem there is exactly one 11613. other option:$V(2.511167869) \approx 49$. It would be like knowing that 11614. the square is king of the silo, sitting in splendid glory on top of your 11615. grain. 11616. 11617. Fred's eyes light up with delight at this elegant resolution, but on 11618. closer inspection of Quentin's drawings and calculations his happiness 11619. turns to despondency once more. Fred points out to Quentin that it's the 11620. radius of the silo that is 6 metres, not the diameter, and the angle of 11621. repose for his grain is 40 degrees. However, if Quentin can find a set of 11622. solutions for this particular silo then he will be more than happy to keep 11623. it. 11624. 11625. If Quick thinking Quentin is to satisfy frustratingly fussy Fred the 11626. farmer's appetite for all things square then determine the values of$x$11627. for all possible square space wastage options and calculate$\sum x\$
11628.    correct to 9 decimal places.
11629.
11630.
11631.    p_431_grain_silo.png
11633.
11634.
11635. Problem 432
11636. ===========
11637.
11638.
11639.    Let S(n,m) = ∑φ(n × i) for 1 ≤ i ≤ m. (φ is Euler's totient function)
11640.    You are given that S(510510,10^6 )= 45480596821125120.
11641.
11642.    Find S(510510,10^11).
11644.
11645.
11647.
11648.
11649. Problem 433
11650. ===========
11651.
11652.
11653.    Let E(x, y) be the number of steps it takes to determine the
11654.    greatest common divisor of x and y with Euclid's algorithm. More
11655.    formally:
11656.    x = y, y = x mod y
11657.    x[n] = y[n-1], y[n] = x[n-1] mod y[n-1]
11658.    E(x, y) is the smallest n such that y[n] = 0.
11659.
11660.    We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4.
11661.
11662.    Define S(N) as the sum of E(x,y) for 1 ≤ x,y ≤ N.
11663.    We have S(1) = 1, S(10) = 221 and S(100) = 39826.
11664.
11665.    Find S(5·10^6).
11666.
11667.
11669.
11670.
11671. Problem 434
11672. ===========
11673.
11674.
11675.    Recall that a graph is a collection of vertices and edges connecting the
11676.    vertices, and that two vertices connected by an edge are called adjacent.
11677.    Graphs can be embedded in Euclidean space by associating each vertex with
11678.    a point in the Euclidean space.
11679.    A flexible graph is an embedding of a graph where it is possible to move
11680.    one or more vertices continuously so that the distance between at least
11681.    two nonadjacent vertices is altered while the distances between each pair
11682.    of adjacent vertices is kept constant.
11683.    A rigid graph is an embedding of a graph which is not flexible.
11684.    Informally, a graph is rigid if by replacing the vertices with fully
11685.    rotating hinges and the edges with rods that are unbending and inelastic,
11686.    no parts of the graph can be moved independently from the rest of the
11687.    graph.
11688.
11689.    The grid graphs embedded in the Euclidean plane are not rigid, as the
11690.    following animation demonstrates:
11691.
11692.    However, one can make them rigid by adding diagonal edges to the cells.
11693.    For example, for the 2x3 grid graph, there are 19 ways to make the graph
11694.    rigid:
11695.
11696.    Note that for the purposes of this problem, we do not consider changing
11697.    the orientation of a diagonal edge or adding both diagonal edges to a cell
11698.    as a different way of making a grid graph rigid.
11699.
11700.    Let R(m,n) be the number of ways to make the m × n grid graph rigid.
11701.    E.g. R(2,3) = 19 and R(5,5) = 23679901
11702.
11703.    Define S(N) as ∑R(i,j) for 1 ≤ i, j ≤ N.
11704.    E.g. S(5) = 25021721.
11706.
11707.
11709.
11710.
11711. Problem 435
11712. ===========
11713.
11714.
11715.    The Fibonacci numbers {f[n], n ≥ 0} are defined recursively as f[n] =
11716.    f[n-1] + f[n-2] with base cases f = 0 and f = 1.
11717.
11718.    Define the polynomials {F[n], n ≥ 0} as F[n](x) = ∑f[i]x^i for 0 ≤ i ≤ n.
11719.
11720.    For example, F(x) = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7, and
11721.    F(11) = 268357683.
11722.
11723.    Let n = 10^15. Find the sum [∑[0≤x≤100] F[n](x)] mod 1307674368000 (=
11724.    15!).
11725.
11726.
11728.
11729.
11730. Problem 436
11731. ===========
11732.
11733.
11734.    Julie proposes the following wager to her sister Louise.
11735.    She suggests they play a game of chance to determine who will wash the
11736.    dishes.
11737.    For this game, they shall use a generator of independent random numbers
11738.    uniformly distributed between 0 and 1.
11739.    The game starts with S = 0.
11740.    The first player, Louise, adds to S different random numbers from the
11741.    generator until S > 1 and records her last random number 'x'.
11742.    The second player, Julie, continues adding to S different random numbers
11743.    from the generator until S > 2 and records her last random number 'y'.
11744.    The player with the highest number wins and the loser washes the dishes,
11745.    i.e. if y > x the second player wins.
11746.
11747.    For example, if the first player draws 0.62 and 0.44, the first player
11748.    turn ends since 0.62+0.44 > 1 and x = 0.44.
11749.    If the second players draws 0.1, 0.27 and 0.91, the second player turn
11750.    ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91.Since y > x, the
11751.    second player wins.
11752.
11753.    Louise thinks about it for a second, and objects: "That's not fair".
11754.    What is the probability that the second player wins?
11755.    Give your answer rounded to 10 places behind the decimal point in the form
11756.    0.abcdefghij
11757.
11758.
11760.
11761.
11762. Problem 437
11763. ===========
11764.
11765.
11766.    When we calculate 8^n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3,
11767.    2, 5, 7.
11768.    As we see all possible values from 1 to 10 occur. So 8 is a primitive root
11769.    of 11.
11770.    But there is more:
11771.    If we take a closer look we see:
11772.    1+8=9
11773.    8+9=17≡6 mod 11
11774.    9+6=15≡4 mod 11
11775.    6+4=10
11776.    4+10=14≡3 mod 11
11777.    10+3=13≡2 mod 11
11778.    3+2=5
11779.    2+5=7
11780.    5+7=12≡1 mod 11.
11781.
11782.    So the powers of 8 mod 11 are cyclic with period 10, and 8^n + 8^n+1 ≡
11783.    8^n+2 (mod 11).
11784.    8 is called a Fibonacci primitive root of 11.
11785.    Not every prime has a Fibonacci primitive root.
11786.    There are 323 primes less than 10000 with one or more Fibonacci primitive
11787.    roots and the sum of these primes is 1480491.
11788.    Find the sum of the primes less than 100,000,000 with at least one
11789.    Fibonacci primitive root.
11790.