Not a member of Pastebin yet?
Sign Up,
it unlocks many cool features!
- Attribution-Non-Commercial-Share Alike 2.0, Attribution to Kyle Keen at kmkeen.com/local-euler/
- Project Euler is protected under
- Attribution-Non-Commercial-Share Alike 2.0 UK: England & Wales
- ^sup and [sub] are frequently use for subscripts/superscripts.
- Many symbols are utf8, my apologies if you are on a 7-bit TTY.
- Solutions are hashed with md5sum
- echo -n 'myanswer' | md5sum
- generated on 2014-6-29 at 15:46
- Problem 1
- =========
- If we list all the natural numbers below 10 that are multiples of 3 or 5,
- we get 3, 5, 6 and 9. The sum of these multiples is 23.
- Find the sum of all the multiples of 3 or 5 below 1000.
- Answer: 233168
- Problem 2
- =========
- Each new term in the Fibonacci sequence is generated by adding the
- previous two terms. By starting with 1 and 2, the first 10 terms will be:
- 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
- By considering the terms in the Fibonacci sequence whose values do not
- exceed four million, find the sum of the even-valued terms.
- Answer: 4613732
- Problem 3
- =========
- The prime factors of 13195 are 5, 7, 13 and 29.
- What is the largest prime factor of the number 600851475143 ?
- Answer: 6857
- Problem 4
- =========
- A palindromic number reads the same both ways. The largest palindrome made
- from the product of two 2-digit numbers is 9009 = 91 × 99.
- Find the largest palindrome made from the product of two 3-digit numbers.
- Answer: 906609
- Problem 5
- =========
- 2520 is the smallest number that can be divided by each of the numbers
- from 1 to 10 without any remainder.
- What is the smallest positive number that is evenly divisible by all of
- the numbers from 1 to 20?
- Answer: 232792560
- Problem 6
- =========
- The sum of the squares of the first ten natural numbers is,
- 1^2 + 2^2 + ... + 10^2 = 385
- The square of the sum of the first ten natural numbers is,
- (1 + 2 + ... + 10)^2 = 55^2 = 3025
- Hence the difference between the sum of the squares of the first ten
- natural numbers and the square of the sum is 3025 − 385 = 2640.
- Find the difference between the sum of the squares of the first one
- hundred natural numbers and the square of the sum.
- Answer: 25164150
- Problem 7
- =========
- By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
- that the 6th prime is 13.
- What is the 10 001st prime number?
- Answer: 104743
- Problem 8
- =========
- The four adjacent digits in the 1000-digit number that have the greatest
- product are 9 × 9 × 8 × 9 = 5832.
- 73167176531330624919225119674426574742355349194934
- 96983520312774506326239578318016984801869478851843
- 85861560789112949495459501737958331952853208805511
- 12540698747158523863050715693290963295227443043557
- 66896648950445244523161731856403098711121722383113
- 62229893423380308135336276614282806444486645238749
- 30358907296290491560440772390713810515859307960866
- 70172427121883998797908792274921901699720888093776
- 65727333001053367881220235421809751254540594752243
- 52584907711670556013604839586446706324415722155397
- 53697817977846174064955149290862569321978468622482
- 83972241375657056057490261407972968652414535100474
- 82166370484403199890008895243450658541227588666881
- 16427171479924442928230863465674813919123162824586
- 17866458359124566529476545682848912883142607690042
- 24219022671055626321111109370544217506941658960408
- 07198403850962455444362981230987879927244284909188
- 84580156166097919133875499200524063689912560717606
- 05886116467109405077541002256983155200055935729725
- 71636269561882670428252483600823257530420752963450
- Find the thirteen adjacent digits in the 1000-digit number that have the
- greatest product. What is the value of this product?
- Answer: 23514624000
- Problem 9
- =========
- A Pythagorean triplet is a set of three natural numbers, a < b < c, for
- which,
- a^2 + b^2 = c^2
- For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
- There exists exactly one Pythagorean triplet for which a + b + c = 1000.
- Find the product abc.
- Answer: 31875000
- Problem 10
- ==========
- The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
- Find the sum of all the primes below two million.
- Answer: 142913828922
- Problem 11
- ==========
- In the 20×20 grid below, four numbers along a diagonal line have been
- marked in red.
- 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
- 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
- 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
- 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
- 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
- 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
- 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
- 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
- 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
- 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
- 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
- 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
- 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
- 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
- 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
- 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
- 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
- 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
- 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
- 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
- The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
- What is the greatest product of four adjacent numbers in the same
- direction (up, down, left, right, or diagonally) in the 20×20 grid?
- Answer: 70600674
- Problem 12
- ==========
- The sequence of triangle numbers is generated by adding the natural
- numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 =
- 28. The first ten terms would be:
- 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
- Let us list the factors of the first seven triangle numbers:
- 1: 1
- 3: 1,3
- 6: 1,2,3,6
- 10: 1,2,5,10
- 15: 1,3,5,15
- 21: 1,3,7,21
- 28: 1,2,4,7,14,28
- We can see that 28 is the first triangle number to have over five
- divisors.
- What is the value of the first triangle number to have over five hundred
- divisors?
- Answer: 76576500
- Problem 13
- ==========
- Work out the first ten digits of the sum of the following one-hundred
- 50-digit numbers.
- 37107287533902102798797998220837590246510135740250
- 46376937677490009712648124896970078050417018260538
- 74324986199524741059474233309513058123726617309629
- 91942213363574161572522430563301811072406154908250
- 23067588207539346171171980310421047513778063246676
- 89261670696623633820136378418383684178734361726757
- 28112879812849979408065481931592621691275889832738
- 44274228917432520321923589422876796487670272189318
- 47451445736001306439091167216856844588711603153276
- 70386486105843025439939619828917593665686757934951
- 62176457141856560629502157223196586755079324193331
- 64906352462741904929101432445813822663347944758178
- 92575867718337217661963751590579239728245598838407
- 58203565325359399008402633568948830189458628227828
- 80181199384826282014278194139940567587151170094390
- 35398664372827112653829987240784473053190104293586
- 86515506006295864861532075273371959191420517255829
- 71693888707715466499115593487603532921714970056938
- 54370070576826684624621495650076471787294438377604
- 53282654108756828443191190634694037855217779295145
- 36123272525000296071075082563815656710885258350721
- 45876576172410976447339110607218265236877223636045
- 17423706905851860660448207621209813287860733969412
- 81142660418086830619328460811191061556940512689692
- 51934325451728388641918047049293215058642563049483
- 62467221648435076201727918039944693004732956340691
- 15732444386908125794514089057706229429197107928209
- 55037687525678773091862540744969844508330393682126
- 18336384825330154686196124348767681297534375946515
- 80386287592878490201521685554828717201219257766954
- 78182833757993103614740356856449095527097864797581
- 16726320100436897842553539920931837441497806860984
- 48403098129077791799088218795327364475675590848030
- 87086987551392711854517078544161852424320693150332
- 59959406895756536782107074926966537676326235447210
- 69793950679652694742597709739166693763042633987085
- 41052684708299085211399427365734116182760315001271
- 65378607361501080857009149939512557028198746004375
- 35829035317434717326932123578154982629742552737307
- 94953759765105305946966067683156574377167401875275
- 88902802571733229619176668713819931811048770190271
- 25267680276078003013678680992525463401061632866526
- 36270218540497705585629946580636237993140746255962
- 24074486908231174977792365466257246923322810917141
- 91430288197103288597806669760892938638285025333403
- 34413065578016127815921815005561868836468420090470
- 23053081172816430487623791969842487255036638784583
- 11487696932154902810424020138335124462181441773470
- 63783299490636259666498587618221225225512486764533
- 67720186971698544312419572409913959008952310058822
- 95548255300263520781532296796249481641953868218774
- 76085327132285723110424803456124867697064507995236
- 37774242535411291684276865538926205024910326572967
- 23701913275725675285653248258265463092207058596522
- 29798860272258331913126375147341994889534765745501
- 18495701454879288984856827726077713721403798879715
- 38298203783031473527721580348144513491373226651381
- 34829543829199918180278916522431027392251122869539
- 40957953066405232632538044100059654939159879593635
- 29746152185502371307642255121183693803580388584903
- 41698116222072977186158236678424689157993532961922
- 62467957194401269043877107275048102390895523597457
- 23189706772547915061505504953922979530901129967519
- 86188088225875314529584099251203829009407770775672
- 11306739708304724483816533873502340845647058077308
- 82959174767140363198008187129011875491310547126581
- 97623331044818386269515456334926366572897563400500
- 42846280183517070527831839425882145521227251250327
- 55121603546981200581762165212827652751691296897789
- 32238195734329339946437501907836945765883352399886
- 75506164965184775180738168837861091527357929701337
- 62177842752192623401942399639168044983993173312731
- 32924185707147349566916674687634660915035914677504
- 99518671430235219628894890102423325116913619626622
- 73267460800591547471830798392868535206946944540724
- 76841822524674417161514036427982273348055556214818
- 97142617910342598647204516893989422179826088076852
- 87783646182799346313767754307809363333018982642090
- 10848802521674670883215120185883543223812876952786
- 71329612474782464538636993009049310363619763878039
- 62184073572399794223406235393808339651327408011116
- 66627891981488087797941876876144230030984490851411
- 60661826293682836764744779239180335110989069790714
- 85786944089552990653640447425576083659976645795096
- 66024396409905389607120198219976047599490197230297
- 64913982680032973156037120041377903785566085089252
- 16730939319872750275468906903707539413042652315011
- 94809377245048795150954100921645863754710598436791
- 78639167021187492431995700641917969777599028300699
- 15368713711936614952811305876380278410754449733078
- 40789923115535562561142322423255033685442488917353
- 44889911501440648020369068063960672322193204149535
- 41503128880339536053299340368006977710650566631954
- 81234880673210146739058568557934581403627822703280
- 82616570773948327592232845941706525094512325230608
- 22918802058777319719839450180888072429661980811197
- 77158542502016545090413245809786882778948721859617
- 72107838435069186155435662884062257473692284509516
- 20849603980134001723930671666823555245252804609722
- 53503534226472524250874054075591789781264330331690
- Answer: 5537376230
- Problem 14
- ==========
- The following iterative sequence is defined for the set of positive
- integers:
- n → n/2 (n is even)
- n → 3n + 1 (n is odd)
- Using the rule above and starting with 13, we generate the following
- sequence:
- 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
- It can be seen that this sequence (starting at 13 and finishing at 1)
- contains 10 terms. Although it has not been proved yet (Collatz Problem),
- it is thought that all starting numbers finish at 1.
- Which starting number, under one million, produces the longest chain?
- NOTE: Once the chain starts the terms are allowed to go above one million.
- Answer: 837799
- Problem 15
- ==========
- Starting in the top left corner of a 2×2 grid, and only being able to move
- to the right and down, there are exactly 6 routes to the bottom right
- corner.
- How many such routes are there through a 20×20 grid?
- p_015.gif
- Answer: 137846528820
- Problem 16
- ==========
- 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
- What is the sum of the digits of the number 2^1000?
- Answer: 1366
- Problem 17
- ==========
- If the numbers 1 to 5 are written out in words: one, two, three, four,
- five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
- If all the numbers from 1 to 1000 (one thousand) inclusive were written
- out in words, how many letters would be used?
- NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
- forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
- 20 letters. The use of "and" when writing out numbers is in compliance
- with British usage.
- Answer: 21124
- Problem 18
- ==========
- By starting at the top of the triangle below and moving to adjacent
- numbers on the row below, the maximum total from top to bottom is 23.
- 3
- 7 4
- 2 4 6
- 8 5 9 3
- That is, 3 + 7 + 4 + 9 = 23.
- Find the maximum total from top to bottom of the triangle below:
- 75
- 95 64
- 17 47 82
- 18 35 87 10
- 20 04 82 47 65
- 19 01 23 75 03 34
- 88 02 77 73 07 63 67
- 99 65 04 28 06 16 70 92
- 41 41 26 56 83 40 80 70 33
- 41 48 72 33 47 32 37 16 94 29
- 53 71 44 65 25 43 91 52 97 51 14
- 70 11 33 28 77 73 17 78 39 68 17 57
- 91 71 52 38 17 14 91 43 58 50 27 29 48
- 63 66 04 68 89 53 67 30 73 16 69 87 40 31
- 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
- NOTE: As there are only 16384 routes, it is possible to solve this problem
- by trying every route. However, [1]Problem 67, is the same challenge with
- a triangle containing one-hundred rows; it cannot be solved by brute
- force, and requires a clever method! ;o)
- Visible links
- 1. problem=67
- Answer: 1074
- Problem 19
- ==========
- You are given the following information, but you may prefer to do some
- research for yourself.
- • 1 Jan 1900 was a Monday.
- • Thirty days has September,
- April, June and November.
- All the rest have thirty-one,
- Saving February alone,
- Which has twenty-eight, rain or shine.
- And on leap years, twenty-nine.
- • A leap year occurs on any year evenly divisible by 4, but not on a
- century unless it is divisible by 400.
- How many Sundays fell on the first of the month during the twentieth
- century (1 Jan 1901 to 31 Dec 2000)?
- Answer: 171
- Problem 20
- ==========
- n! means n × (n − 1) × ... × 3 × 2 × 1
- For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
- and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 =
- 27.
- Find the sum of the digits in the number 100!
- Answer: 648
- Problem 21
- ==========
- Let d(n) be defined as the sum of proper divisors of n (numbers less than
- n which divide evenly into n).
- If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair
- and each of a and b are called amicable numbers.
- For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
- 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1,
- 2, 4, 71 and 142; so d(284) = 220.
- Evaluate the sum of all the amicable numbers under 10000.
- Answer: 31626
- Problem 22
- ==========
- Using [1]names.txt, a 46K text file containing over five-thousand first
- names, begin by sorting it into alphabetical order. Then working out the
- alphabetical value for each name, multiply this value by its alphabetical
- position in the list to obtain a name score.
- For example, when the list is sorted into alphabetical order, COLIN, which
- is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So,
- COLIN would obtain a score of 938 × 53 = 49714.
- What is the total of all the name scores in the file?
- Visible links
- 1. names.txt
- Answer: 871198282
- Problem 23
- ==========
- A perfect number is a number for which the sum of its proper divisors is
- exactly equal to the number. For example, the sum of the proper divisors
- of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
- number.
- A number n is called deficient if the sum of its proper divisors is less
- than n and it is called abundant if this sum exceeds n.
- As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the
- smallest number that can be written as the sum of two abundant numbers is
- 24. By mathematical analysis, it can be shown that all integers greater
- than 28123 can be written as the sum of two abundant numbers. However,
- this upper limit cannot be reduced any further by analysis even though it
- is known that the greatest number that cannot be expressed as the sum of
- two abundant numbers is less than this limit.
- Find the sum of all the positive integers which cannot be written as the
- sum of two abundant numbers.
- Answer: 4179871
- Problem 24
- ==========
- A permutation is an ordered arrangement of objects. For example, 3124 is
- one possible permutation of the digits 1, 2, 3 and 4. If all of the
- permutations are listed numerically or alphabetically, we call it
- lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
- 012 021 102 120 201 210
- What is the millionth lexicographic permutation of the digits 0, 1, 2, 3,
- 4, 5, 6, 7, 8 and 9?
- Answer: 7f155b45cb3f0a6e518d59ec348bff84
- Problem 25
- ==========
- The Fibonacci sequence is defined by the recurrence relation:
- F[n] = F[n−1] + F[n−2], where F[1] = 1 and F[2] = 1.
- Hence the first 12 terms will be:
- F[1] = 1
- F[2] = 1
- F[3] = 2
- F[4] = 3
- F[5] = 5
- F[6] = 8
- F[7] = 13
- F[8] = 21
- F[9] = 34
- F[10] = 55
- F[11] = 89
- F[12] = 144
- The 12th term, F[12], is the first term to contain three digits.
- What is the first term in the Fibonacci sequence to contain 1000 digits?
- Answer: 4782
- Problem 26
- ==========
- A unit fraction contains 1 in the numerator. The decimal representation of
- the unit fractions with denominators 2 to 10 are given:
- 1/2 = 0.5
- 1/3 = 0.(3)
- 1/4 = 0.25
- 1/5 = 0.2
- 1/6 = 0.1(6)
- 1/7 = 0.(142857)
- 1/8 = 0.125
- 1/9 = 0.(1)
- 1/10 = 0.1
- Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can
- be seen that 1/7 has a 6-digit recurring cycle.
- Find the value of d < 1000 for which ^1/[d] contains the longest recurring
- cycle in its decimal fraction part.
- Answer: 983
- Problem 27
- ==========
- Euler discovered the remarkable quadratic formula:
- n² + n + 41
- It turns out that the formula will produce 40 primes for the consecutive
- values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41
- is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly
- divisible by 41.
- The incredible formula n² − 79n + 1601 was discovered, which produces 80
- primes for the consecutive values n = 0 to 79. The product of the
- coefficients, −79 and 1601, is −126479.
- Considering quadratics of the form:
- n² + an + b, where |a| < 1000 and |b| < 1000
- where |n| is the modulus/absolute value of n
- e.g. |11| = 11 and |−4| = 4
- Find the product of the coefficients, a and b, for the quadratic
- expression that produces the maximum number of primes for consecutive
- values of n, starting with n = 0.
- Answer: 69d9e3218fd7abb6ff453ea96505183d
- Problem 28
- ==========
- Starting with the number 1 and moving to the right in a clockwise
- direction a 5 by 5 spiral is formed as follows:
- 21 22 23 24 25
- 20 7 8 9 10
- 19 6 1 2 11
- 18 5 4 3 12
- 17 16 15 14 13
- It can be verified that the sum of the numbers on the diagonals is 101.
- What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral
- formed in the same way?
- Answer: 0d53425bd7c5bf9919df3718c8e49fa6
- Problem 29
- ==========
- Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
- 2^2=4, 2^3=8, 2^4=16, 2^5=32
- 3^2=9, 3^3=27, 3^4=81, 3^5=243
- 4^2=16, 4^3=64, 4^4=256, 4^5=1024
- 5^2=25, 5^3=125, 5^4=625, 5^5=3125
- If they are then placed in numerical order, with any repeats removed, we
- get the following sequence of 15 distinct terms:
- 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
- How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤
- 100 and 2 ≤ b ≤ 100?
- Answer: 9183
- Problem 30
- ==========
- Surprisingly there are only three numbers that can be written as the sum
- of fourth powers of their digits:
- 1634 = 1^4 + 6^4 + 3^4 + 4^4
- 8208 = 8^4 + 2^4 + 0^4 + 8^4
- 9474 = 9^4 + 4^4 + 7^4 + 4^4
- As 1 = 1^4 is not a sum it is not included.
- The sum of these numbers is 1634 + 8208 + 9474 = 19316.
- Find the sum of all the numbers that can be written as the sum of fifth
- powers of their digits.
- Answer: 443839
- Problem 31
- ==========
- In England the currency is made up of pound, £, and pence, p, and there
- are eight coins in general circulation:
- 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
- It is possible to make £2 in the following way:
- 1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
- How many different ways can £2 be made using any number of coins?
- Answer: 73682
- Problem 32
- ==========
- We shall say that an n-digit number is pandigital if it makes use of all
- the digits 1 to n exactly once; for example, the 5-digit number, 15234, is
- 1 through 5 pandigital.
- The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
- multiplicand, multiplier, and product is 1 through 9 pandigital.
- Find the sum of all products whose multiplicand/multiplier/product
- identity can be written as a 1 through 9 pandigital.
- HINT: Some products can be obtained in more than one way so be sure to
- only include it once in your sum.
- Answer: 45228
- Problem 33
- ==========
- The fraction 49/98 is a curious fraction, as an inexperienced
- mathematician in attempting to simplify it may incorrectly believe that
- 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
- We shall consider fractions like, 30/50 = 3/5, to be trivial
- examples.
- There are exactly four non-trivial examples of this type of fraction, less
- than one in value, and containing two digits in the numerator and
- denominator.
- If the product of these four fractions is given in its lowest common
- terms, find the value of the denominator.
- Answer: 100
- Problem 34
- ==========
- 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
- Find the sum of all numbers which are equal to the sum of the factorial of
- their digits.
- Note: as 1! = 1 and 2! = 2 are not sums they are not included.
- Answer: 40730
- Problem 35
- ==========
- The number, 197, is called a circular prime because all rotations of the
- digits: 197, 971, and 719, are themselves prime.
- There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37,
- 71, 73, 79, and 97.
- How many circular primes are there below one million?
- Answer: 55
- Problem 36
- ==========
- The decimal number, 585 = 1001001001[2] (binary), is palindromic in both
- bases.
- Find the sum of all numbers, less than one million, which are palindromic
- in base 10 and base 2.
- (Please note that the palindromic number, in either base, may not include
- leading zeros.)
- Answer: 872187
- Problem 37
- ==========
- The number 3797 has an interesting property. Being prime itself, it is
- possible to continuously remove digits from left to right, and remain
- prime at each stage: 3797, 797, 97, and 7. Similarly we can work from
- right to left: 3797, 379, 37, and 3.
- Find the sum of the only eleven primes that are both truncatable from left
- to right and right to left.
- NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
- Answer: 748317
- Problem 38
- ==========
- Take the number 192 and multiply it by each of 1, 2, and 3:
- 192 × 1 = 192
- 192 × 2 = 384
- 192 × 3 = 576
- By concatenating each product we get the 1 to 9 pandigital, 192384576. We
- will call 192384576 the concatenated product of 192 and (1,2,3)
- The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4,
- and 5, giving the pandigital, 918273645, which is the concatenated product
- of 9 and (1,2,3,4,5).
- What is the largest 1 to 9 pandigital 9-digit number that can be formed as
- the concatenated product of an integer with (1,2, ... , n) where n > 1?
- Answer: f2a29ede8dc9fae7926dc7a4357ac25e
- Problem 39
- ==========
- If p is the perimeter of a right angle triangle with integral length
- sides, {a,b,c}, there are exactly three solutions for p = 120.
- {20,48,52}, {24,45,51}, {30,40,50}
- For which value of p ≤ 1000, is the number of solutions maximised?
- Answer: 840
- Problem 40
- ==========
- An irrational decimal fraction is created by concatenating the positive
- integers:
- 0.123456789101112131415161718192021...
- It can be seen that the 12^th digit of the fractional part is 1.
- If d[n] represents the n^th digit of the fractional part, find the value
- of the following expression.
- d[1] × d[10] × d[100] × d[1000] × d[10000] × d[100000] × d[1000000]
- Answer: 210
- Problem 41
- ==========
- We shall say that an n-digit number is pandigital if it makes use of all
- the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital
- and is also prime.
- What is the largest n-digit pandigital prime that exists?
- Answer: 7652413
- Problem 42
- ==========
- The n^th term of the sequence of triangle numbers is given by, t[n] =
- ½n(n+1); so the first ten triangle numbers are:
- 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
- By converting each letter in a word to a number corresponding to its
- alphabetical position and adding these values we form a word value. For
- example, the word value for SKY is 19 + 11 + 25 = 55 = t[10]. If the word
- value is a triangle number then we shall call the word a triangle word.
- Using [1]words.txt, a 16K text file containing nearly two-thousand common
- English words, how many are triangle words?
- Visible links
- 1. words.txt
- Answer: 162
- Problem 43
- ==========
- The number, 1406357289, is a 0 to 9 pandigital number because it is made
- up of each of the digits 0 to 9 in some order, but it also has a rather
- interesting sub-string divisibility property.
- Let d[1] be the 1^st digit, d[2] be the 2^nd digit, and so on. In this
- way, we note the following:
- • d[2]d[3]d[4]=406 is divisible by 2
- • d[3]d[4]d[5]=063 is divisible by 3
- • d[4]d[5]d[6]=635 is divisible by 5
- • d[5]d[6]d[7]=357 is divisible by 7
- • d[6]d[7]d[8]=572 is divisible by 11
- • d[7]d[8]d[9]=728 is divisible by 13
- • d[8]d[9]d[10]=289 is divisible by 17
- Find the sum of all 0 to 9 pandigital numbers with this property.
- Answer: 115253b7721af0fdff25cd391dfc70cf
- Problem 44
- ==========
- Pentagonal numbers are generated by the formula, P[n]=n(3n−1)/2. The first
- ten pentagonal numbers are:
- 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
- It can be seen that P[4] + P[7] = 22 + 70 = 92 = P[8]. However, their
- difference, 70 − 22 = 48, is not pentagonal.
- Find the pair of pentagonal numbers, P[j] and P[k], for which their sum
- and difference are pentagonal and D = |P[k] − P[j]| is minimised; what is
- the value of D?
- Answer: 5482660
- Problem 45
- ==========
- Triangle, pentagonal, and hexagonal numbers are generated by the following
- formulae:
- Triangle T[n]=n(n+1)/2 1, 3, 6, 10, 15, ...
- Pentagonal P[n]=n(3n−1)/2 1, 5, 12, 22, 35, ...
- Hexagonal H[n]=n(2n−1) 1, 6, 15, 28, 45, ...
- It can be verified that T[285] = P[165] = H[143] = 40755.
- Find the next triangle number that is also pentagonal and hexagonal.
- Answer: 30dfe3e3b286add9d12e493ca7be63fc
- Problem 46
- ==========
- It was proposed by Christian Goldbach that every odd composite number can
- be written as the sum of a prime and twice a square.
- 9 = 7 + 2×1^2
- 15 = 7 + 2×2^2
- 21 = 3 + 2×3^2
- 25 = 7 + 2×3^2
- 27 = 19 + 2×2^2
- 33 = 31 + 2×1^2
- It turns out that the conjecture was false.
- What is the smallest odd composite that cannot be written as the sum of a
- prime and twice a square?
- Answer: 5777
- Problem 47
- ==========
- The first two consecutive numbers to have two distinct prime factors are:
- 14 = 2 × 7
- 15 = 3 × 5
- The first three consecutive numbers to have three distinct prime factors
- are:
- 644 = 2² × 7 × 23
- 645 = 3 × 5 × 43
- 646 = 2 × 17 × 19.
- Find the first four consecutive integers to have four distinct prime
- factors. What is the first of these numbers?
- Answer: 134043
- Problem 48
- ==========
- The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
- Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
- Answer: 0829124724747ae1c65da8cae5263346
- Problem 49
- ==========
- The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
- increases by 3330, is unusual in two ways: (i) each of the three terms are
- prime, and, (ii) each of the 4-digit numbers are permutations of one
- another.
- There are no arithmetic sequences made up of three 1-, 2-, or 3-digit
- primes, exhibiting this property, but there is one other 4-digit
- increasing sequence.
- What 12-digit number do you form by concatenating the three terms in this
- sequence?
- Answer: 0b99933d3e2a9addccbb663d46cbb592
- Problem 50
- ==========
- The prime 41, can be written as the sum of six consecutive primes:
- 41 = 2 + 3 + 5 + 7 + 11 + 13
- This is the longest sum of consecutive primes that adds to a prime below
- one-hundred.
- The longest sum of consecutive primes below one-thousand that adds to a
- prime, contains 21 terms, and is equal to 953.
- Which prime, below one-million, can be written as the sum of the most
- consecutive primes?
- Answer: 997651
- Problem 51
- ==========
- By replacing the 1^st digit of the 2-digit number *3, it turns out that
- six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all
- prime.
- By replacing the 3^rd and 4^th digits of 56**3 with the same digit, this
- 5-digit number is the first example having seven primes among the ten
- generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663,
- 56773, and 56993. Consequently 56003, being the first member of this
- family, is the smallest prime with this property.
- Find the smallest prime which, by replacing part of the number (not
- necessarily adjacent digits) with the same digit, is part of an eight
- prime value family.
- Answer: 121313
- Problem 52
- ==========
- It can be seen that the number, 125874, and its double, 251748, contain
- exactly the same digits, but in a different order.
- Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x,
- contain the same digits.
- Answer: 142857
- Problem 53
- ==========
- There are exactly ten ways of selecting three from five, 12345:
- 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
- In combinatorics, we use the notation, ^5C[3] = 10.
- In general,
- ^nC[r] = n! ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
- r!(n−r)!
- It is not until n = 23, that a value exceeds one-million: ^23C[10] =
- 1144066.
- How many, not necessarily distinct, values of ^nC[r], for 1 ≤ n ≤ 100,
- are greater than one-million?
- Answer: 4075
- Problem 54
- ==========
- In the card game poker, a hand consists of five cards and are ranked, from
- lowest to highest, in the following way:
- • High Card: Highest value card.
- • One Pair: Two cards of the same value.
- • Two Pairs: Two different pairs.
- • Three of a Kind: Three cards of the same value.
- • Straight: All cards are consecutive values.
- • Flush: All cards of the same suit.
- • Full House: Three of a kind and a pair.
- • Four of a Kind: Four cards of the same value.
- • Straight Flush: All cards are consecutive values of same suit.
- • Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
- The cards are valued in the order:
- 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
- If two players have the same ranked hands then the rank made up of the
- highest value wins; for example, a pair of eights beats a pair of fives
- (see example 1 below). But if two ranks tie, for example, both players
- have a pair of queens, then highest cards in each hand are compared (see
- example 4 below); if the highest cards tie then the next highest cards are
- compared, and so on.
- Consider the following five hands dealt to two players:
- Hand Player 1 Player 2 Winner
- 1 5H 5C 6S 7S KD 2C 3S 8S 8D TD Player 2
- Pair of Fives Pair of Eights
- 2 5D 8C 9S JS AC 2C 5C 7D 8S QH Player 1
- Highest card Ace Highest card Queen
- 3 2D 9C AS AH AC 3D 6D 7D TD QD Player 2
- Three Aces Flush with Diamonds
- 4D 6S 9H QH QC 3D 6D 7H QD QS
- 4 Pair of Queens Pair of Queens Player 1
- Highest card Nine Highest card Seven
- 2H 2D 4C 4D 4S 3C 3D 3S 9S 9D
- 5 Full House Full House Player 1
- With Three Fours with Three Threes
- The file, [1]poker.txt, contains one-thousand random hands dealt to two
- players. Each line of the file contains ten cards (separated by a single
- space): the first five are Player 1's cards and the last five are Player
- 2's cards. You can assume that all hands are valid (no invalid characters
- or repeated cards), each player's hand is in no specific order, and in
- each hand there is a clear winner.
- How many hands does Player 1 win?
- Visible links
- 1. poker.txt
- Answer: 376
- Problem 55
- ==========
- If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
- Not all numbers produce palindromes so quickly. For example,
- 349 + 943 = 1292,
- 1292 + 2921 = 4213
- 4213 + 3124 = 7337
- That is, 349 took three iterations to arrive at a palindrome.
- Although no one has proved it yet, it is thought that some numbers, like
- 196, never produce a palindrome. A number that never forms a palindrome
- through the reverse and add process is called a Lychrel number. Due to the
- theoretical nature of these numbers, and for the purpose of this problem,
- we shall assume that a number is Lychrel until proven otherwise. In
- addition you are given that for every number below ten-thousand, it will
- either (i) become a palindrome in less than fifty iterations, or, (ii) no
- one, with all the computing power that exists, has managed so far to map
- it to a palindrome. In fact, 10677 is the first number to be shown to
- require over fifty iterations before producing a palindrome:
- 4668731596684224866951378664 (53 iterations, 28-digits).
- Surprisingly, there are palindromic numbers that are themselves Lychrel
- numbers; the first example is 4994.
- How many Lychrel numbers are there below ten-thousand?
- NOTE: Wording was modified slightly on 24 April 2007 to emphasise the
- theoretical nature of Lychrel numbers.
- Answer: 249
- Problem 56
- ==========
- A googol (10^100) is a massive number: one followed by one-hundred zeros;
- 100^100 is almost unimaginably large: one followed by two-hundred zeros.
- Despite their size, the sum of the digits in each number is only 1.
- Considering natural numbers of the form, a^b, where a, b < 100, what is
- the maximum digital sum?
- Answer: 972
- Problem 57
- ==========
- It is possible to show that the square root of two can be expressed as an
- infinite continued fraction.
- √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
- By expanding this for the first four iterations, we get:
- 1 + 1/2 = 3/2 = 1.5
- 1 + 1/(2 + 1/2) = 7/5 = 1.4
- 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
- 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
- The next three expansions are 99/70, 239/169, and 577/408, but the eighth
- expansion, 1393/985, is the first example where the number of digits in
- the numerator exceeds the number of digits in the denominator.
- In the first one-thousand expansions, how many fractions contain a
- numerator with more digits than denominator?
- Answer: 153
- Problem 58
- ==========
- Starting with 1 and spiralling anticlockwise in the following way, a
- square spiral with side length 7 is formed.
- 37 36 35 34 33 32 31
- 38 17 16 15 14 13 30
- 39 18 5 4 3 12 29
- 40 19 6 1 2 11 28
- 41 20 7 8 9 10 27
- 42 21 22 23 24 25 26
- 43 44 45 46 47 48 49
- It is interesting to note that the odd squares lie along the bottom right
- diagonal, but what is more interesting is that 8 out of the 13 numbers
- lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
- If one complete new layer is wrapped around the spiral above, a square
- spiral with side length 9 will be formed. If this process is continued,
- what is the side length of the square spiral for which the ratio of primes
- along both diagonals first falls below 10%?
- Answer: 26241
- Problem 59
- ==========
- Each character on a computer is assigned a unique code and the preferred
- standard is ASCII (American Standard Code for Information Interchange).
- For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
- A modern encryption method is to take a text file, convert the bytes to
- ASCII, then XOR each byte with a given value, taken from a secret key. The
- advantage with the XOR function is that using the same encryption key on
- the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
- then 107 XOR 42 = 65.
- For unbreakable encryption, the key is the same length as the plain text
- message, and the key is made up of random bytes. The user would keep the
- encrypted message and the encryption key in different locations, and
- without both "halves", it is impossible to decrypt the message.
- Unfortunately, this method is impractical for most users, so the modified
- method is to use a password as a key. If the password is shorter than the
- message, which is likely, the key is repeated cyclically throughout the
- message. The balance for this method is using a sufficiently long password
- key for security, but short enough to be memorable.
- Your task has been made easy, as the encryption key consists of three
- lower case characters. Using [1]cipher1.txt, a file containing the
- encrypted ASCII codes, and the knowledge that the plain text must contain
- common English words, decrypt the message and find the sum of the ASCII
- values in the original text.
- Visible links
- 1. cipher1.txt
- Answer: 107359
- Problem 60
- ==========
- The primes 3, 7, 109, and 673, are quite remarkable. By taking any two
- primes and concatenating them in any order the result will always be
- prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The
- sum of these four primes, 792, represents the lowest sum for a set of four
- primes with this property.
- Find the lowest sum for a set of five primes for which any two primes
- concatenate to produce another prime.
- Answer: 26033
- Problem 61
- ==========
- Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers
- are all figurate (polygonal) numbers and are generated by the following
- formulae:
- Triangle P[3,n]=n(n+1)/2 1, 3, 6, 10, 15, ...
- Square P[4,n]=n^2 1, 4, 9, 16, 25, ...
- Pentagonal P[5,n]=n(3n−1)/2 1, 5, 12, 22, 35, ...
- Hexagonal P[6,n]=n(2n−1) 1, 6, 15, 28, 45, ...
- Heptagonal P[7,n]=n(5n−3)/2 1, 7, 18, 34, 55, ...
- Octagonal P[8,n]=n(3n−2) 1, 8, 21, 40, 65, ...
- The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three
- interesting properties.
- 1. The set is cyclic, in that the last two digits of each number is the
- first two digits of the next number (including the last number with
- the first).
- 2. Each polygonal type: triangle (P[3,127]=8128), square (P[4,91]=8281),
- and pentagonal (P[5,44]=2882), is represented by a different number in
- the set.
- 3. This is the only set of 4-digit numbers with this property.
- Find the sum of the only ordered set of six cyclic 4-digit numbers for
- which each polygonal type: triangle, square, pentagonal, hexagonal,
- heptagonal, and octagonal, is represented by a different number in the
- set.
- Answer: 28684
- Problem 62
- ==========
- The cube, 41063625 (345^3), can be permuted to produce two other cubes:
- 56623104 (384^3) and 66430125 (405^3). In fact, 41063625 is the smallest
- cube which has exactly three permutations of its digits which are also
- cube.
- Find the smallest cube for which exactly five permutations of its digits
- are cube.
- Answer: 8f46b522b5401b8b6df99a7410eea44b
- Problem 63
- ==========
- The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the
- 9-digit number, 134217728=8^9, is a ninth power.
- How many n-digit positive integers exist which are also an nth power?
- Answer: 49
- Problem 64
- ==========
- All square roots are periodic when written as continued fractions and can
- be written in the form:
- √N = a[0] + 1
- a[1] + 1
- a[2] + 1
- a[3] + ...
- For example, let us consider √23:
- √23 = 4 + √23 — 4 = 4 + 1 = 4 + 1
- 1 1 + √23 – 3
- √23—4 7
- If we continue we would get the following expansion:
- √23 = 4 + 1
- 1 + 1
- 3 + 1
- 1 + 1
- 8 + ...
- The process can be summarised as follows:
- a[0] = 4, 1 = √23+4 = 1 + √23—3
- √23—4 7 7
- a[1] = 1, 7 = 7(√23+3) = 3 + √23—3
- √23—3 14 2
- a[2] = 3, 2 = 2(√23+3) = 1 + √23—4
- √23—3 14 7
- a[3] = 1, 7 = 7(√23+4) = 8 + √23—4
- √23—4 7
- a[4] = 8, 1 = √23+4 = 1 + √23—3
- √23—4 7 7
- a[5] = 1, 7 = 7(√23+3) = 3 + √23—3
- √23—3 14 2
- a[6] = 3, 2 = 2(√23+3) = 1 + √23—4
- √23—3 14 7
- a[7] = 1, 7 = 7(√23+4) = 8 + √23—4
- √23—4 7
- It can be seen that the sequence is repeating. For conciseness, we use the
- notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats
- indefinitely.
- The first ten continued fraction representations of (irrational) square
- roots are:
- √2=[1;(2)], period=1
- √3=[1;(1,2)], period=2
- √5=[2;(4)], period=1
- √6=[2;(2,4)], period=2
- √7=[2;(1,1,1,4)], period=4
- √8=[2;(1,4)], period=2
- √10=[3;(6)], period=1
- √11=[3;(3,6)], period=2
- √12= [3;(2,6)], period=2
- √13=[3;(1,1,1,1,6)], period=5
- Exactly four continued fractions, for N ≤ 13, have an odd period.
- How many continued fractions for N ≤ 10000 have an odd period?
- Answer: 1322
- Problem 65
- ==========
- The square root of 2 can be written as an infinite continued fraction.
- √2 = 1 + 1
- 2 + 1
- 2 + 1
- 2 + 1
- 2 + ...
- The infinite continued fraction can be written, √2 = [1;(2)], (2)
- indicates that 2 repeats ad infinitum. In a similar way, √23 =
- [4;(1,3,1,8)].
- It turns out that the sequence of partial values of continued fractions
- for square roots provide the best rational approximations. Let us consider
- the convergents for √2.
- 1 + 1 = 3/2
- 2
- 1 + 1 = 7/5
- 2 + 1
- 2
- 1 + 1 = 17/12
- 2 + 1
- 2 + 1
- 2
- 1 + 1 = 41/29
- 2 + 1
- 2 + 1
- 2 + 1
- 2
- Hence the sequence of the first ten convergents for √2 are:
- 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378,
- ...
- What is most surprising is that the important mathematical constant,
- e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
- The first ten terms in the sequence of convergents for e are:
- 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
- The sum of digits in the numerator of the 10^th convergent is 1+4+5+7=17.
- Find the sum of digits in the numerator of the 100^th convergent of the
- continued fraction for e.
- Answer: 272
- Problem 66
- ==========
- Consider quadratic Diophantine equations of the form:
- x^2 – Dy^2 = 1
- For example, when D=13, the minimal solution in x is 649^2 – 13×180^2 = 1.
- It can be assumed that there are no solutions in positive integers when D
- is square.
- By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the
- following:
- 3^2 – 2×2^2 = 1
- 2^2 – 3×1^2 = 1
- 9^2 – 5×4^2 = 1
- 5^2 – 6×2^2 = 1
- 8^2 – 7×3^2 = 1
- Hence, by considering minimal solutions in x for D ≤ 7, the largest x is
- obtained when D=5.
- Find the value of D ≤ 1000 in minimal solutions of x for which the largest
- value of x is obtained.
- Answer: 661
- Problem 67
- ==========
- By starting at the top of the triangle below and moving to adjacent
- numbers on the row below, the maximum total from top to bottom is 23.
- 3
- 7 4
- 2 4 6
- 8 5 9 3
- That is, 3 + 7 + 4 + 9 = 23.
- Find the maximum total from top to bottom in [1]triangle.txt, a 15K text
- file containing a triangle with one-hundred rows.
- NOTE: This is a much more difficult version of [2]Problem 18. It is not
- possible to try every route to solve this problem, as there are 2^99
- altogether! If you could check one trillion (10^12) routes every second it
- would take over twenty billion years to check them all. There is an
- efficient algorithm to solve it. ;o)
- Visible links
- 1. triangle.txt
- 2. problem=18
- Answer: 7273
- Problem 68
- ==========
- Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6,
- and each line adding to nine.
- Working clockwise, and starting from the group of three with the
- numerically lowest external node (4,3,2 in this example), each solution
- can be described uniquely. For example, the above solution can be
- described by the set: 4,3,2; 6,2,1; 5,1,3.
- It is possible to complete the ring with four different totals: 9, 10, 11,
- and 12. There are eight solutions in total.
- Total Solution Set
- 9 4,2,3; 5,3,1; 6,1,2
- 9 4,3,2; 6,2,1; 5,1,3
- 10 2,3,5; 4,5,1; 6,1,3
- 10 2,5,3; 6,3,1; 4,1,5
- 11 1,4,6; 3,6,2; 5,2,4
- 11 1,6,4; 5,4,2; 3,2,6
- 12 1,5,6; 2,6,4; 3,4,5
- 12 1,6,5; 3,5,4; 2,4,6
- By concatenating each group it is possible to form 9-digit strings; the
- maximum string for a 3-gon ring is 432621513.
- Using the numbers 1 to 10, and depending on arrangements, it is possible
- to form 16- and 17-digit strings. What is the maximum 16-digit string for
- a "magic" 5-gon ring?
- p_068_1.gif
- p_068_2.gif
- Answer: 26227442c6fed0292a528ac3790175be
- Problem 69
- ==========
- Euler's Totient function, φ(n) [sometimes called the phi function], is
- used to determine the number of numbers less than n which are relatively
- prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine
- and relatively prime to nine, φ(9)=6.
- ┌────┬──────────────────┬──────┬───────────┐
- │ n │ Relatively Prime │ φ(n) │ n/φ(n) │
- ├────┼──────────────────┼──────┼───────────┤
- │ 2 │ 1 │ 1 │ 2 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 3 │ 1,2 │ 2 │ 1.5 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 4 │ 1,3 │ 2 │ 2 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 5 │ 1,2,3,4 │ 4 │ 1.25 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 6 │ 1,5 │ 2 │ 3 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 7 │ 1,2,3,4,5,6 │ 6 │ 1.1666... │
- ├────┼──────────────────┼──────┼───────────┤
- │ 8 │ 1,3,5,7 │ 4 │ 2 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 9 │ 1,2,4,5,7,8 │ 6 │ 1.5 │
- ├────┼──────────────────┼──────┼───────────┤
- │ 10 │ 1,3,7,9 │ 4 │ 2.5 │
- └────┴──────────────────┴──────┴───────────┘
- It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
- Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
- Answer: 510510
- Problem 70
- ==========
- Euler's Totient function, φ(n) [sometimes called the phi function], is
- used to determine the number of positive numbers less than or equal to n
- which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are
- all less than nine and relatively prime to nine, φ(9)=6.
- The number 1 is considered to be relatively prime to every positive
- number, so φ(1)=1.
- Interestingly, φ(87109)=79180, and it can be seen that 87109 is a
- permutation of 79180.
- Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n
- and the ratio n/φ(n) produces a minimum.
- Answer: 8319823
- Problem 71
- ==========
- Consider the fraction, n/d, where n and d are positive integers. If n<d
- and HCF(n,d)=1, it is called a reduced proper fraction.
- If we list the set of reduced proper fractions for d ≤ 8 in ascending
- order of size, we get:
- 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,
- 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
- It can be seen that 2/5 is the fraction immediately to the left of 3/7.
- By listing the set of reduced proper fractions for d ≤ 1,000,000 in
- ascending order of size, find the numerator of the fraction immediately to
- the left of 3/7.
- Answer: 428570
- Problem 72
- ==========
- Consider the fraction, n/d, where n and d are positive integers. If n<d
- and HCF(n,d)=1, it is called a reduced proper fraction.
- If we list the set of reduced proper fractions for d ≤ 8 in ascending
- order of size, we get:
- 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,
- 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
- It can be seen that there are 21 elements in this set.
- How many elements would be contained in the set of reduced proper
- fractions for d ≤ 1,000,000?
- Answer: 0384fb529dc651fe0f460acff3e9ac5d
- Problem 73
- ==========
- Consider the fraction, n/d, where n and d are positive integers. If n<d
- and HCF(n,d)=1, it is called a reduced proper fraction.
- If we list the set of reduced proper fractions for d ≤ 8 in ascending
- order of size, we get:
- 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,
- 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
- It can be seen that there are 3 fractions between 1/3 and 1/2.
- How many fractions lie between 1/3 and 1/2 in the sorted set of reduced
- proper fractions for d ≤ 12,000?
- Answer: 7295372
- Problem 74
- ==========
- The number 145 is well known for the property that the sum of the
- factorial of its digits is equal to 145:
- 1! + 4! + 5! = 1 + 24 + 120 = 145
- Perhaps less well known is 169, in that it produces the longest chain of
- numbers that link back to 169; it turns out that there are only three such
- loops that exist:
- 169 → 363601 → 1454 → 169
- 871 → 45361 → 871
- 872 → 45362 → 872
- It is not difficult to prove that EVERY starting number will eventually
- get stuck in a loop. For example,
- 69 → 363600 → 1454 → 169 → 363601 (→ 1454)
- 78 → 45360 → 871 → 45361 (→ 871)
- 540 → 145 (→ 145)
- Starting with 69 produces a chain of five non-repeating terms, but the
- longest non-repeating chain with a starting number below one million is
- sixty terms.
- How many chains, with a starting number below one million, contain exactly
- sixty non-repeating terms?
- Answer: 402
- Problem 75
- ==========
- It turns out that 12 cm is the smallest length of wire that can be bent to
- form an integer sided right angle triangle in exactly one way, but there
- are many more examples.
- 12 cm: (3,4,5)
- 24 cm: (6,8,10)
- 30 cm: (5,12,13)
- 36 cm: (9,12,15)
- 40 cm: (8,15,17)
- 48 cm: (12,16,20)
- In contrast, some lengths of wire, like 20 cm, cannot be bent to form an
- integer sided right angle triangle, and other lengths allow more than one
- solution to be found; for example, using 120 cm it is possible to form
- exactly three different integer sided right angle triangles.
- 120 cm: (30,40,50), (20,48,52), (24,45,51)
- Given that L is the length of the wire, for how many values of L ≤
- 1,500,000 can exactly one integer sided right angle triangle be formed?
- Answer: 161667
- Problem 76
- ==========
- It is possible to write five as a sum in exactly six different ways:
- 4 + 1
- 3 + 2
- 3 + 1 + 1
- 2 + 2 + 1
- 2 + 1 + 1 + 1
- 1 + 1 + 1 + 1 + 1
- How many different ways can one hundred be written as a sum of at least
- two positive integers?
- Answer: 190569291
- Problem 77
- ==========
- It is possible to write ten as the sum of primes in exactly five different
- ways:
- 7 + 3
- 5 + 5
- 5 + 3 + 2
- 3 + 3 + 2 + 2
- 2 + 2 + 2 + 2 + 2
- What is the first value which can be written as the sum of primes in over
- five thousand different ways?
- Answer: 71
- Problem 78
- ==========
- Let p(n) represent the number of different ways in which n coins can be
- separated into piles. For example, five coins can separated into piles in
- exactly seven different ways, so p(5)=7.
- OOOOO
- OOOO O
- OOO OO
- OOO O O
- OO OO O
- OO O O O
- O O O O O
- Find the least value of n for which p(n) is divisible by one million.
- Answer: 55374
- Problem 79
- ==========
- A common security method used for online banking is to ask the user for
- three random characters from a passcode. For example, if the passcode was
- 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected
- reply would be: 317.
- The text file, [1]keylog.txt, contains fifty successful login attempts.
- Given that the three characters are always asked for in order, analyse the
- file so as to determine the shortest possible secret passcode of unknown
- length.
- Visible links
- 1. keylog.txt
- Answer: 73162890
- Problem 80
- ==========
- It is well known that if the square root of a natural number is not an
- integer, then it is irrational. The decimal expansion of such square roots
- is infinite without any repeating pattern at all.
- The square root of two is 1.41421356237309504880..., and the digital sum
- of the first one hundred decimal digits is 475.
- For the first one hundred natural numbers, find the total of the digital
- sums of the first one hundred decimal digits for all the irrational square
- roots.
- Answer: 40886
- Problem 81
- ==========
- In the 5 by 5 matrix below, the minimal path sum from the top left to the
- bottom right, by only moving to the right and down, is indicated in bold
- red and is equal to 2427.
- 131 673 234 103 18
- 201 96 342 965 150
- 630 803 746 422 111
- 537 699 497 121 956
- 805 732 524 37 331
- Find the minimal path sum, in [1]matrix.txt, a 31K text file containing a
- 80 by 80 matrix, from the top left to the bottom right by only moving
- right and down.
- Visible links
- 1. matrix.txt
- Answer: 427337
- Problem 82
- ==========
- NOTE: This problem is a more challenging version of [1]Problem 81.
- The minimal path sum in the 5 by 5 matrix below, by starting in any cell
- in the left column and finishing in any cell in the right column, and only
- moving up, down, and right, is indicated in red and bold; the sum is equal
- to 994.
- 131 673 234 103 18
- 201 96 342 965 150
- 630 803 746 422 111
- 537 699 497 121 956
- 805 732 524 37 331
- Find the minimal path sum, in [2]matrix.txt, a 31K text file containing a
- 80 by 80 matrix, from the left column to the right column.
- Visible links
- 1. problem=81
- 2. matrix.txt
- Answer: 260324
- Problem 83
- ==========
- NOTE: This problem is a significantly more challenging version of
- [1]Problem 81.
- In the 5 by 5 matrix below, the minimal path sum from the top left to the
- bottom right, by moving left, right, up, and down, is indicated in bold
- red and is equal to 2297.
- 131 673 234 103 18
- 201 96 342 965 150
- 630 803 746 422 111
- 537 699 497 121 956
- 805 732 524 37 331
- Find the minimal path sum, in [2]matrix.txt, a 31K text file containing a
- 80 by 80 matrix, from the top left to the bottom right by moving left,
- right, up, and down.
- Visible links
- 1. problem=81
- 2. matrix.txt
- Answer: 425185
- Problem 84
- ==========
- In the game, Monopoly, the standard board is set up in the following way:
- GO A1 CC1 A2 T1 R1 B1 CH1 B2 B3 JAIL
- H2 C1
- T2 U1
- H1 C2
- CH3 C3
- R4 R2
- G3 D1
- CC3 CC2
- G2 D2
- G1 D3
- G2J F3 U2 F2 F1 R3 E3 E2 CH2 E1 FP
- A player starts on the GO square and adds the scores on two 6-sided dice
- to determine the number of squares they advance in a clockwise direction.
- Without any further rules we would expect to visit each square with equal
- probability: 2.5%. However, landing on G2J (Go To Jail), CC (community
- chest), and CH (chance) changes this distribution.
- In addition to G2J, and one card from each of CC and CH, that orders the
- player to go directly to jail, if a player rolls three consecutive
- doubles, they do not advance the result of their 3rd roll. Instead they
- proceed directly to jail.
- At the beginning of the game, the CC and CH cards are shuffled. When a
- player lands on CC or CH they take a card from the top of the respective
- pile and, after following the instructions, it is returned to the bottom
- of the pile. There are sixteen cards in each pile, but for the purpose of
- this problem we are only concerned with cards that order a movement; any
- instruction not concerned with movement will be ignored and the player
- will remain on the CC/CH square.
- • Community Chest (2/16 cards):
- 1. Advance to GO
- 2. Go to JAIL
- • Chance (10/16 cards):
- 1. Advance to GO
- 2. Go to JAIL
- 3. Go to C1
- 4. Go to E3
- 5. Go to H2
- 6. Go to R1
- 7. Go to next R (railway company)
- 8. Go to next R
- 9. Go to next U (utility company)
- 10. Go back 3 squares.
- The heart of this problem concerns the likelihood of visiting a particular
- square. That is, the probability of finishing at that square after a roll.
- For this reason it should be clear that, with the exception of G2J for
- which the probability of finishing on it is zero, the CH squares will have
- the lowest probabilities, as 5/8 request a movement to another square, and
- it is the final square that the player finishes at on each roll that we
- are interested in. We shall make no distinction between "Just Visiting"
- and being sent to JAIL, and we shall also ignore the rule about requiring
- a double to "get out of jail", assuming that they pay to get out on their
- next turn.
- By starting at GO and numbering the squares sequentially from 00 to 39 we
- can concatenate these two-digit numbers to produce strings that correspond
- with sets of squares.
- Statistically it can be shown that the three most popular squares, in
- order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO
- (3.09%) = Square 00. So these three most popular squares can be listed
- with the six-digit modal string: 102400.
- If, instead of using two 6-sided dice, two 4-sided dice are used, find the
- six-digit modal string.
- Answer: 101524
- Problem 85
- ==========
- By counting carefully it can be seen that a rectangular grid measuring 3
- by 2 contains eighteen rectangles:
- Although there exists no rectangular grid that contains exactly two
- million rectangles, find the area of the grid with the nearest solution.
- p_085.gif
- Answer: 2772
- Problem 86
- ==========
- A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3,
- and a fly, F, sits in the opposite corner. By travelling on the surfaces
- of the room the shortest "straight line" distance from S to F is 10 and
- the path is shown on the diagram.
- However, there are up to three "shortest" path candidates for any given
- cuboid and the shortest route doesn't always have integer length.
- By considering all cuboid rooms with integer dimensions, up to a maximum
- size of M by M by M, there are exactly 2060 cuboids for which the shortest
- route has integer length when M=100, and this is the least value of M for
- which the number of solutions first exceeds two thousand; the number of
- solutions is 1975 when M=99.
- Find the least value of M such that the number of solutions first exceeds
- one million.
- p_086.gif
- Answer: 1818
- Problem 87
- ==========
- The smallest number expressible as the sum of a prime square, prime cube,
- and prime fourth power is 28. In fact, there are exactly four numbers
- below fifty that can be expressed in such a way:
- 28 = 2^2 + 2^3 + 2^4
- 33 = 3^2 + 2^3 + 2^4
- 49 = 5^2 + 2^3 + 2^4
- 47 = 2^2 + 3^3 + 2^4
- How many numbers below fifty million can be expressed as the sum of a
- prime square, prime cube, and prime fourth power?
- Answer: 1097343
- Problem 88
- ==========
- A natural number, N, that can be written as the sum and product of a given
- set of at least two natural numbers, {a[1], a[2], ... , a[k]} is called a
- product-sum number: N = a[1] + a[2] + ... + a[k] = a[1] × a[2] × ... ×
- a[k].
- For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
- For a given set of size, k, we shall call the smallest N with this
- property a minimal product-sum number. The minimal product-sum numbers for
- sets of size, k = 2, 3, 4, 5, and 6 are as follows.
- k=2: 4 = 2 × 2 = 2 + 2
- k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
- k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
- k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
- k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
- Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is
- 4+6+8+12 = 30; note that 8 is only counted once in the sum.
- In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is
- {4, 6, 8, 12, 15, 16}, the sum is 61.
- What is the sum of all the minimal product-sum numbers for 2≤k≤12000?
- Answer: 7587457
- Problem 89
- ==========
- The rules for writing Roman numerals allow for many ways of writing each
- number (see [1]About Roman Numerals...). However, there is always a "best"
- way of writing a particular number.
- For example, the following represent all of the legitimate ways of writing
- the number sixteen:
- IIIIIIIIIIIIIIII
- VIIIIIIIIIII
- VVIIIIII
- XIIIIII
- VVVI
- XVI
- The last example being considered the most efficient, as it uses the least
- number of numerals.
- The 11K text file, [2]roman.txt, contains one thousand numbers written in
- valid, but not necessarily minimal, Roman numerals; that is, they are
- arranged in descending units and obey the subtractive pair rule (see
- [3]About Roman Numerals... for the definitive rules for this problem).
- Find the number of characters saved by writing each of these in their
- minimal form.
- Note: You can assume that all the Roman numerals in the file contain no
- more than four consecutive identical units.
- Visible links
- 1. about=roman_numerals
- 2. roman.txt
- 3. about=roman_numerals
- Answer: 743
- Problem 90
- ==========
- Each of the six faces on a cube has a different digit (0 to 9) written on
- it; the same is done to a second cube. By placing the two cubes
- side-by-side in different positions we can form a variety of 2-digit
- numbers.
- For example, the square number 64 could be formed:
- In fact, by carefully choosing the digits on both cubes it is possible to
- display all of the square numbers below one-hundred: 01, 04, 09, 16, 25,
- 36, 49, 64, and 81.
- For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9}
- on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
- However, for this problem we shall allow the 6 or 9 to be turned
- upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3,
- 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it
- would be impossible to obtain 09.
- In determining a distinct arrangement we are interested in the digits on
- each cube, not the order.
- {1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
- {1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
- But because we are allowing 6 and 9 to be reversed, the two distinct sets
- in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9}
- for the purpose of forming 2-digit numbers.
- How many distinct arrangements of the two cubes allow for all of the
- square numbers to be displayed?
- p_090.gif
- Answer: 1217
- Problem 91
- ==========
- The points P (x[1], y[1]) and Q (x[2], y[2]) are plotted at integer
- co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.
- There are exactly fourteen triangles containing a right angle that can be
- formed when each co-ordinate lies between 0 and 2 inclusive; that is,
- 0 ≤ x[1], y[1], x[2], y[2] ≤ 2.
- Given that 0 ≤ x[1], y[1], x[2], y[2] ≤ 50, how many right triangles can
- be formed?
- p_091_1.gif
- p_091_2.gif
- Answer: 14234
- Problem 92
- ==========
- A number chain is created by continuously adding the square of the digits
- in a number to form a new number until it has been seen before.
- For example,
- 44 → 32 → 13 → 10 → 1 → 1
- 85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89
- Therefore any chain that arrives at 1 or 89 will become stuck in an
- endless loop. What is most amazing is that EVERY starting number will
- eventually arrive at 1 or 89.
- How many starting numbers below ten million will arrive at 89?
- Answer: 8581146
- Problem 93
- ==========
- By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and
- making use of the four arithmetic operations (+, −, *, /) and
- brackets/parentheses, it is possible to form different positive integer
- targets.
- For example,
- 8 = (4 * (1 + 3)) / 2
- 14 = 4 * (3 + 1 / 2)
- 19 = 4 * (2 + 3) − 1
- 36 = 3 * 4 * (2 + 1)
- Note that concatenations of the digits, like 12 + 34, are not allowed.
- Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different
- target numbers of which 36 is the maximum, and each of the numbers 1 to 28
- can be obtained before encountering the first non-expressible number.
- Find the set of four distinct digits, a < b < c < d, for which the longest
- set of consecutive positive integers, 1 to n, can be obtained, giving your
- answer as a string: abcd.
- Answer: 1258
- Problem 94
- ==========
- It is easily proved that no equilateral triangle exists with integral
- length sides and integral area. However, the almost equilateral triangle
- 5-5-6 has an area of 12 square units.
- We shall define an almost equilateral triangle to be a triangle for which
- two sides are equal and the third differs by no more than one unit.
- Find the sum of the perimeters of all almost equilateral triangles with
- integral side lengths and area and whose perimeters do not exceed one
- billion (1,000,000,000).
- Answer: 3218c6bb59f2539ec39ad4bf37c10913
- Problem 95
- ==========
- The proper divisors of a number are all the divisors excluding the number
- itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As
- the sum of these divisors is equal to 28, we call it a perfect number.
- Interestingly the sum of the proper divisors of 220 is 284 and the sum of
- the proper divisors of 284 is 220, forming a chain of two numbers. For
- this reason, 220 and 284 are called an amicable pair.
- Perhaps less well known are longer chains. For example, starting with
- 12496, we form a chain of five numbers:
- 12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)
- Since this chain returns to its starting point, it is called an amicable
- chain.
- Find the smallest member of the longest amicable chain with no element
- exceeding one million.
- Answer: 14316
- Problem 96
- ==========
- Su Doku (Japanese meaning number place) is the name given to a popular
- puzzle concept. Its origin is unclear, but credit must be attributed to
- Leonhard Euler who invented a similar, and much more difficult, puzzle
- idea called Latin Squares. The objective of Su Doku puzzles, however, is
- to replace the blanks (or zeros) in a 9 by 9 grid in such that each row,
- column, and 3 by 3 box contains each of the digits 1 to 9. Below is an
- example of a typical starting puzzle grid and its solution grid.
- ┌───────┬───────┬───────┐ ┌───────┬───────┬───────┐
- │ 0 0 3 │ 0 2 0 │ 6 0 0 │ │ 4 8 3 │ 9 2 1 │ 6 5 7 │
- │ 9 0 0 │ 3 0 5 │ 0 0 1 │ │ 9 6 7 │ 3 4 5 │ 8 2 1 │
- │ 0 0 1 │ 8 0 6 │ 4 0 0 │ │ 2 5 1 │ 8 7 6 │ 4 9 3 │
- ├───────┼───────┼───────┤ ├───────┼───────┼───────┤
- │ 0 0 8 │ 1 0 2 │ 9 0 0 │ │ 5 4 8 │ 1 3 2 │ 9 7 6 │
- │ 7 0 0 │ 0 0 0 │ 0 0 8 │ │ 7 2 9 │ 5 6 4 │ 1 3 8 │
- │ 0 0 6 │ 7 0 8 │ 2 0 0 │ │ 1 3 6 │ 7 9 8 │ 2 4 5 │
- ├───────┼───────┼───────┤ ├───────┼───────┼───────┤
- │ 0 0 2 │ 6 0 9 │ 5 0 0 │ │ 3 7 2 │ 6 8 9 │ 5 1 4 │
- │ 8 0 0 │ 2 0 3 │ 0 0 9 │ │ 8 1 4 │ 2 5 3 │ 7 6 9 │
- │ 0 0 5 │ 0 1 0 │ 3 0 0 │ │ 6 9 5 │ 4 1 7 │ 3 8 2 │
- └───────┴───────┴───────┘ └───────┴───────┴───────┘
- A well constructed Su Doku puzzle has a unique solution and can be solved
- by logic, although it may be necessary to employ "guess and test" methods
- in order to eliminate options (there is much contested opinion over this).
- The complexity of the search determines the difficulty of the puzzle; the
- example above is considered easy because it can be solved by straight
- forward direct deduction.
- The 6K text file, [1]sudoku.txt, contains fifty different Su Doku puzzles
- ranging in difficulty, but all with unique solutions (the first puzzle in
- the file is the example above).
- By solving all fifty puzzles find the sum of the 3-digit numbers found in
- the top left corner of each solution grid; for example, 483 is the 3-digit
- number found in the top left corner of the solution grid above.
- Visible links
- 1. sudoku.txt
- Answer: 24702
- Problem 97
- ==========
- The first known prime found to exceed one million digits was discovered in
- 1999, and is a Mersenne prime of the form 2^6972593−1; it contains exactly
- 2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p−1,
- have been found which contain more digits.
- However, in 2004 there was found a massive non-Mersenne prime which
- contains 2,357,207 digits: 28433×2^7830457+1.
- Find the last ten digits of this prime number.
- Answer: 68c8c919526039022b923a72d5cc12b1
- Problem 98
- ==========
- By replacing each of the letters in the word CARE with 1, 2, 9, and 6
- respectively, we form a square number: 1296 = 36^2. What is remarkable is
- that, by using the same digital substitutions, the anagram, RACE, also
- forms a square number: 9216 = 96^2. We shall call CARE (and RACE) a square
- anagram word pair and specify further that leading zeroes are not
- permitted, neither may a different letter have the same digital value as
- another letter.
- Using [1]words.txt, a 16K text file containing nearly two-thousand common
- English words, find all the square anagram word pairs (a palindromic word
- is NOT considered to be an anagram of itself).
- What is the largest square number formed by any member of such a pair?
- NOTE: All anagrams formed must be contained in the given text file.
- Visible links
- 1. words.txt
- Answer: 18769
- Problem 99
- ==========
- Comparing two numbers written in index form like 2^11 and 3^7 is not
- difficult, as any calculator would confirm that 2^11 = 2048 < 3^7 = 2187.
- However, confirming that 632382^518061 > 519432^525806 would be much more
- difficult, as both numbers contain over three million digits.
- Using [1]base_exp.txt, a 22K text file containing one thousand lines with
- a base/exponent pair on each line, determine which line number has the
- greatest numerical value.
- NOTE: The first two lines in the file represent the numbers in the example
- given above.
- Visible links
- 1. base_exp.txt
- Answer: 709
- Problem 100
- ===========
- If a box contains twenty-one coloured discs, composed of fifteen blue
- discs and six red discs, and two discs were taken at random, it can be
- seen that the probability of taking two blue discs, P(BB) =
- (15/21)×(14/20) = 1/2.
- The next such arrangement, for which there is exactly 50% chance of taking
- two blue discs at random, is a box containing eighty-five blue discs and
- thirty-five red discs.
- By finding the first arrangement to contain over 10^12 = 1,000,000,000,000
- discs in total, determine the number of blue discs that the box would
- contain.
- Answer: 21156e3acc4ca35b7a318c541a0648d5
- Problem 101
- ===========
- If we are presented with the first k terms of a sequence it is impossible
- to say with certainty the value of the next term, as there are infinitely
- many polynomial functions that can model the sequence.
- As an example, let us consider the sequence of cube numbers. This is
- defined by the generating function,
- u[n] = n^3: 1, 8, 27, 64, 125, 216, ...
- Suppose we were only given the first two terms of this sequence. Working
- on the principle that "simple is best" we should assume a linear
- relationship and predict the next term to be 15 (common difference 7).
- Even if we were presented with the first three terms, by the same
- principle of simplicity, a quadratic relationship should be assumed.
- We shall define OP(k, n) to be the n^th term of the optimum polynomial
- generating function for the first k terms of a sequence. It should be
- clear that OP(k, n) will accurately generate the terms of the sequence for
- n ≤ k, and potentially the first incorrect term (FIT) will be OP(k, k+1);
- in which case we shall call it a bad OP (BOP).
- As a basis, if we were only given the first term of sequence, it would be
- most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u[1].
- Hence we obtain the following OPs for the cubic sequence:
- OP(1, n) = 1 1, 1, 1, 1, ...
- OP(2, n) = 7n−6 1, 8, 15, ...
- OP(3, n) = 6n^2−11n+6 1, 8, 27, 58, ...
- OP(4, n) = n^3 1, 8, 27, 64, 125, ...
- Clearly no BOPs exist for k ≥ 4.
- By considering the sum of FITs generated by the BOPs (indicated in red
- above), we obtain 1 + 15 + 58 = 74.
- Consider the following tenth degree polynomial generating function:
- u[n] = 1 − n + n^2 − n^3 + n^4 − n^5 + n^6 − n^7 + n^8 − n^9 + n^10
- Find the sum of FITs for the BOPs.
- Answer: d382b0cc25e82446da83d3a792e1cd27
- Problem 102
- ===========
- Three distinct points are plotted at random on a Cartesian plane, for
- which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.
- Consider the following two triangles:
- A(-340,495), B(-153,-910), C(835,-947)
- X(-175,41), Y(-421,-714), Z(574,-645)
- It can be verified that triangle ABC contains the origin, whereas triangle
- XYZ does not.
- Using [1]triangles.txt, a 27K text file containing the co-ordinates of one
- thousand "random" triangles, find the number of triangles for which the
- interior contains the origin.
- NOTE: The first two examples in the file represent the triangles in the
- example given above.
- Visible links
- 1. triangles.txt
- Answer: 228
- Problem 103
- ===========
- Let S(A) represent the sum of elements in set A of size n. We shall call
- it a special sum set if for any two non-empty disjoint subsets, B and C,
- the following properties are true:
- i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
- ii. If B contains more elements than C then S(B) > S(C).
- If S(A) is minimised for a given n, we shall call it an optimum special
- sum set. The first five optimum special sum sets are given below.
- n = 1: {1}
- n = 2: {1, 2}
- n = 3: {2, 3, 4}
- n = 4: {3, 5, 6, 7}
- n = 5: {6, 9, 11, 12, 13}
- It seems that for a given optimum set, A = {a[1], a[2], ... , a[n]}, the
- next optimum set is of the form B = {b, a[1]+b, a[2]+b, ... ,a[n]+b},
- where b is the "middle" element on the previous row.
- By applying this "rule" we would expect the optimum set for n = 6 to be A
- = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the
- optimum set, as we have merely applied an algorithm to provide a near
- optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25},
- with S(A) = 115 and corresponding set string: 111819202225.
- Given that A is an optimum special sum set for n = 7, find its set string.
- NOTE: This problem is related to [1]Problem 105 and [2]Problem 106.
- Visible links
- 1. problem=105
- 2. problem=106
- Answer: af8c238336c2a79bb81a24b3fef3330d
- Problem 104
- ===========
- The Fibonacci sequence is defined by the recurrence relation:
- F[n] = F[n−1] + F[n−2], where F[1] = 1 and F[2] = 1.
- It turns out that F[541], which contains 113 digits, is the first
- Fibonacci number for which the last nine digits are 1-9 pandigital
- (contain all the digits 1 to 9, but not necessarily in order). And
- F[2749], which contains 575 digits, is the first Fibonacci number for
- which the first nine digits are 1-9 pandigital.
- Given that F[k] is the first Fibonacci number for which the first nine
- digits AND the last nine digits are 1-9 pandigital, find k.
- Answer: 329468
- Problem 105
- ===========
- Let S(A) represent the sum of elements in set A of size n. We shall call
- it a special sum set if for any two non-empty disjoint subsets, B and C,
- the following properties are true:
- i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
- ii. If B contains more elements than C then S(B) > S(C).
- For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set
- because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159,
- 161, 139, 158} satisfies both rules for all possible subset pair
- combinations and S(A) = 1286.
- Using [1]sets.txt (right click and "Save Link/Target As..."), a 4K text
- file with one-hundred sets containing seven to twelve elements (the two
- examples given above are the first two sets in the file), identify all the
- special sum sets, A[1], A[2], ..., A[k], and find the value of S(A[1]) +
- S(A[2]) + ... + S(A[k]).
- NOTE: This problem is related to [2]Problem 103 and [3]Problem 106.
- Visible links
- 1. sets.txt
- 2. problem=103
- 3. problem=106
- Answer: 73702
- Problem 106
- ===========
- Let S(A) represent the sum of elements in set A of size n. We shall call
- it a special sum set if for any two non-empty disjoint subsets, B and C,
- the following properties are true:
- i. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
- ii. If B contains more elements than C then S(B) > S(C).
- For this problem we shall assume that a given set contains n strictly
- increasing elements and it already satisfies the second rule.
- Surprisingly, out of the 25 possible subset pairs that can be obtained
- from a set for which n = 4, only 1 of these pairs need to be tested for
- equality (first rule). Similarly, when n = 7, only 70 out of the 966
- subset pairs need to be tested.
- For n = 12, how many of the 261625 subset pairs that can be obtained need
- to be tested for equality?
- NOTE: This problem is related to [1]Problem 103 and [2]Problem 105.
- Visible links
- 1. problem=103
- 2. problem=105
- Answer: 21384
- Problem 107
- ===========
- The following undirected network consists of seven vertices and twelve
- edges with a total weight of 243.
- The same network can be represented by the matrix below.
- ┌──────┬────┬────┬────┬────┬────┬────┬────┐
- │ │ A │ B │ C │ D │ E │ F │ G │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ A │ - │ 16 │ 12 │ 21 │ - │ - │ - │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ B │ 16 │ - │ - │ 17 │ 20 │ - │ - │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ C │ 12 │ - │ - │ 28 │ - │ 31 │ - │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ D │ 21 │ 17 │ 28 │ - │ 18 │ 19 │ 23 │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ E │ - │ 20 │ - │ 18 │ - │ - │ 11 │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ F │ - │ - │ 31 │ 19 │ - │ - │ 27 │
- ├──────┼────┼────┼────┼────┼────┼────┼────┤
- │ G │ - │ - │ - │ 23 │ 11 │ 27 │ - │
- └──────┴────┴────┴────┴────┴────┴────┴────┘
- However, it is possible to optimise the network by removing some edges and
- still ensure that all points on the network remain connected. The network
- which achieves the maximum saving is shown below. It has a weight of 93,
- representing a saving of 243 − 93 = 150 from the original network.
- Using [1]network.txt, a 6K text file containing a network with forty
- vertices, and given in matrix form, find the maximum saving which can be
- achieved by removing redundant edges whilst ensuring that the network
- remains connected.
- Visible links
- 1. network.txt
- p_107_1.gif
- p_107_2.gif
- Answer: 259679
- Problem 108
- ===========
- In the following equation x, y, and n are positive integers.
- 1 1 1
- ─ + ─ = ─
- x y n
- For n = 4 there are exactly three distinct solutions:
- 1 1 1
- ─ + ─ = ─
- 5 20 4
- 1 1 1
- ─ + ─ = ─
- 6 12 4
- 1 1 1
- ─ + ─ = ─
- 8 8 4
- What is the least value of n for which the number of distinct solutions
- exceeds one-thousand?
- NOTE: This problem is an easier version of [1]Problem 110; it is strongly
- advised that you solve this one first.
- Visible links
- 1. problem=110
- Answer: 180180
- Problem 109
- ===========
- In the game of darts a player throws three darts at a target board which
- is split into twenty equal sized sections numbered one to twenty.
- The score of a dart is determined by the number of the region that the
- dart lands in. A dart landing outside the red/green outer ring scores
- zero. The black and cream regions inside this ring represent single
- scores. However, the red/green outer ring and middle ring score double and
- treble scores respectively.
- At the centre of the board are two concentric circles called the bull
- region, or bulls-eye. The outer bull is worth 25 points and the inner bull
- is a double, worth 50 points.
- There are many variations of rules but in the most popular game the
- players will begin with a score 301 or 501 and the first player to reduce
- their running total to zero is a winner. However, it is normal to play a
- "doubles out" system, which means that the player must land a double
- (including the double bulls-eye at the centre of the board) on their final
- dart to win; any other dart that would reduce their running total to one
- or lower means the score for that set of three darts is "bust".
- When a player is able to finish on their current score it is called a
- "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s
- and double bull).
- There are exactly eleven distinct ways to checkout on a score of 6:
- ┌──┬──┬──┐
- │D3│ │ │
- ├──┼──┼──┤
- │D1│D2│ │
- ├──┼──┼──┤
- │S2│D2│ │
- ├──┼──┼──┤
- │D2│D1│ │
- ├──┼──┼──┤
- │S4│D1│ │
- ├──┼──┼──┤
- │S1│S1│D2│
- ├──┼──┼──┤
- │S1│T1│D1│
- ├──┼──┼──┤
- │S1│S3│D1│
- ├──┼──┼──┤
- │D1│D1│D1│
- ├──┼──┼──┤
- │D1│S2│D1│
- ├──┼──┼──┤
- │S2│S2│D1│
- └──┴──┴──┘
- Note that D1 D2 is considered different to D2 D1 as they finish on
- different doubles. However, the combination S1 T1 D1 is considered the
- same as T1 S1 D1.
- In addition we shall not include misses in considering combinations; for
- example, D3 is the same as 0 D3 and 0 0 D3.
- Incredibly there are 42336 distinct ways of checking out in total.
- How many distinct ways can a player checkout with a score less than 100?
- p_109.gif
- Answer: 38182
- Problem 110
- ===========
- In the following equation x, y, and n are positive integers.
- 1 1 1
- ─ + ─ = ─
- x y n
- It can be verified that when n = 1260 there are 113 distinct solutions and
- this is the least value of n for which the total number of distinct
- solutions exceeds one hundred.
- What is the least value of n for which the number of distinct solutions
- exceeds four million?
- NOTE: This problem is a much more difficult version of [1]Problem 108 and
- as it is well beyond the limitations of a brute force approach it requires
- a clever implementation.
- Visible links
- 1. problem=108
- Answer: 591a7a92f10322866e6a02f3b2386a1c
- Problem 111
- ===========
- Considering 4-digit primes containing repeated digits it is clear that
- they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by
- 22, and so on. But there are nine 4-digit primes containing three ones:
- 1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
- We shall say that M(n, d) represents the maximum number of repeated digits
- for an n-digit prime where d is the repeated digit, N(n, d) represents the
- number of such primes, and S(n, d) represents the sum of these primes.
- So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit
- prime where one is the repeated digit, there are N(4, 1) = 9 such primes,
- and the sum of these primes is S(4, 1) = 22275. It turns out that for d =
- 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are
- N(4, 0) = 13 such cases.
- In the same way we obtain the following results for 4-digit primes.
- ┌──────────┬─────────┬─────────┬─────────┐
- │ Digit, d │ M(4, d) │ N(4, d) │ S(4, d) │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 0 │ 2 │ 13 │ 67061 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 1 │ 3 │ 9 │ 22275 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 2 │ 3 │ 1 │ 2221 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 3 │ 3 │ 12 │ 46214 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 4 │ 3 │ 2 │ 8888 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 5 │ 3 │ 1 │ 5557 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 6 │ 3 │ 1 │ 6661 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 7 │ 3 │ 9 │ 57863 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 8 │ 3 │ 1 │ 8887 │
- ├──────────┼─────────┼─────────┼─────────┤
- │ 9 │ 3 │ 7 │ 48073 │
- └──────────┴─────────┴─────────┴─────────┘
- For d = 0 to 9, the sum of all S(4, d) is 273700.
- Find the sum of all S(10, d).
- Answer: cdf4d134a3b0caa10a69e2771ac4fd36
- Problem 112
- ===========
- Working from left-to-right if no digit is exceeded by the digit to its
- left it is called an increasing number; for example, 134468.
- Similarly if no digit is exceeded by the digit to its right it is called a
- decreasing number; for example, 66420.
- We shall call a positive integer that is neither increasing nor decreasing
- a "bouncy" number; for example, 155349.
- Clearly there cannot be any bouncy numbers below one-hundred, but just
- over half of the numbers below one-thousand (525) are bouncy. In fact, the
- least number for which the proportion of bouncy numbers first reaches 50%
- is 538.
- Surprisingly, bouncy numbers become more and more common and by the time
- we reach 21780 the proportion of bouncy numbers is equal to 90%.
- Find the least number for which the proportion of bouncy numbers is
- exactly 99%.
- Answer: 1587000
- Problem 113
- ===========
- Working from left-to-right if no digit is exceeded by the digit to its
- left it is called an increasing number; for example, 134468.
- Similarly if no digit is exceeded by the digit to its right it is called a
- decreasing number; for example, 66420.
- We shall call a positive integer that is neither increasing nor decreasing
- a "bouncy" number; for example, 155349.
- As n increases, the proportion of bouncy numbers below n increases such
- that there are only 12951 numbers below one-million that are not bouncy
- and only 277032 non-bouncy numbers below 10^10.
- How many numbers below a googol (10^100) are not bouncy?
- Answer: a9e504ee704c87f9bddad6d3ffe39532
- Problem 114
- ===========
- A row measuring seven units in length has red blocks with a minimum length
- of three units placed on it, such that any two red blocks (which are
- allowed to be different lengths) are separated by at least one black
- square. There are exactly seventeen ways of doing this.
- ┌┬┬┬┬┬┬┐ ┌──┬┬┬┬┐ ┌┬──┬┬┬┐
- └┴┴┴┴┴┴┘ └──┴┴┴┴┘ └┴──┴┴┴┘
- ┌┬┬──┬┬┐ ┌┬┬┬──┬┐ ┌┬┬┬┬──┐
- └┴┴──┴┴┘ └┴┴┴──┴┘ └┴┴┴┴──┘
- ┌──┬┬──┐ ┌───┬┬┬┐ ┌┬───┬┬┐
- └──┴┴──┘ └───┴┴┴┘ └┴───┴┴┘
- ┌┬┬───┬┐ ┌┬┬┬───┐ ┌────┬┬┐
- └┴┴───┴┘ └┴┴┴───┘ └────┴┴┘
- ┌┬────┬┐ ┌┬┬────┐ ┌─────┬┐
- └┴────┴┘ └┴┴────┘ └─────┴┘
- ┌┬─────┐ ┌──────┐
- └┴─────┘ └──────┘
- How many ways can a row measuring fifty units in length be filled?
- NOTE: Although the example above does not lend itself to the possibility,
- in general it is permitted to mix block sizes. For example, on a row
- measuring eight units in length you could use red (3), black (1), and red
- (4).
- Answer: de48ca72bf252a8be7e0aad762eadcf8
- Problem 115
- ===========
- NOTE: This is a more difficult version of [1]Problem 114.
- A row measuring n units in length has red blocks with a minimum length of
- m units placed on it, such that any two red blocks (which are allowed to
- be different lengths) are separated by at least one black square.
- Let the fill-count function, F(m, n), represent the number of ways that a
- row can be filled.
- For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
- That is, for m = 3, it can be seen that n = 30 is the smallest value for
- which the fill-count function first exceeds one million.
- In the same way, for m = 10, it can be verified that F(10, 56) = 880711
- and F(10, 57) = 1148904, so n = 57 is the least value for which the
- fill-count function first exceeds one million.
- For m = 50, find the least value of n for which the fill-count function
- first exceeds one million.
- Visible links
- 1. problem=114
- Answer: 168
- Problem 116
- ===========
- A row of five black square tiles is to have a number of its tiles replaced
- with coloured oblong tiles chosen from red (length two), green (length
- three), or blue (length four).
- If red tiles are chosen there are exactly seven ways this can be done.
- ┌─╥╥╥┐ ┌╥─╥╥┐ ┌╥╥─╥┐ ┌╥╥╥─┐
- └─╨╨╨┘ └╨─╨╨┘ └╨╨─╨┘ └╨╨╨─┘
- ┌─╥─╥┐ ┌─╥╥─┐ ┌╥─╥─┐
- └─╨─╨┘ └─╨╨─┘ └╨─╨─┘
- If green tiles are chosen there are three ways.
- ┌──╥╥┐ ┌╥──╥┐ ┌╥╥──┐
- └──╨╨┘ └╨──╨┘ └╨╨──┘
- And if blue tiles are chosen there are two ways.
- ┌╥───┐ ┌───╥┐
- └╨───┘ └───╨┘
- Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of
- replacing the black tiles in a row measuring five units in length.
- How many different ways can the black tiles in a row measuring fifty units
- in length be replaced if colours cannot be mixed and at least one coloured
- tile must be used?
- NOTE: This is related to [1]Problem 117.
- Visible links
- 1. problem=117
- Answer: c21ca0ec54e6d1646a953a480f68feb4
- Problem 117
- ===========
- Using a combination of black square tiles and oblong tiles chosen from:
- red tiles measuring two units, green tiles measuring three units, and blue
- tiles measuring four units, it is possible to tile a row measuring five
- units in length in exactly fifteen different ways.
- ┌╥╥╥╥┐ ┌─╥╥╥┐ ┌╥─╥╥┐ ┌╥╥─╥┐
- └╨╨╨╨┘ └─╨╨╨┘ └╨─╨╨┘ └╨╨─╨┘
- ┌╥╥╥─┐ ┌─╥─╥┐ ┌─╥╥─┐ ┌╥─╥─┐
- └╨╨╨─┘ └─╨─╨┘ └─╨╨─┘ └╨─╨─┘
- ┌──╥╥┐ ┌╥──╥┐ ┌╥╥──┐ ┌─╥──┐
- └──╨╨┘ └╨──╨┘ └╨╨──┘ └─╨──┘
- ┌──╥─┐ ┌───╥┐ ┌╥───┐
- └──╨─┘ └───╨┘ └╨───┘
- How many ways can a row measuring fifty units in length be tiled?
- NOTE: This is related to [1]Problem 116.
- Visible links
- 1. problem=116
- Answer: 542612809b3dd08cf518b85450fce8d6
- Problem 118
- ===========
- Using all of the digits 1 through 9 and concatenating them freely to form
- decimal integers, different sets can be formed. Interestingly with the set
- {2,5,47,89,631}, all of the elements belonging to it are prime.
- How many distinct sets containing each of the digits one through nine
- exactly once contain only prime elements?
- Answer: 44680
- Problem 119
- ===========
- The number 512 is interesting because it is equal to the sum of its digits
- raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a
- number with this property is 614656 = 28^4.
- We shall define a[n] to be the nth term of this sequence and insist that a
- number must contain at least two digits to have a sum.
- You are given that a[2] = 512 and a[10] = 614656.
- Find a[30].
- Answer: 72fddfa6c52a120892ade628f3819da4
- Problem 120
- ===========
- Let r be the remainder when (a−1)^n + (a+1)^n is divided by a^2.
- For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49.
- And as n varies, so too will r, but for a = 7 it turns out that r[max] =
- 42.
- For 3 ≤ a ≤ 1000, find ∑ r[max].
- Answer: 0dd05ec40fe11279c2203b72e92a450a
- Problem 121
- ===========
- A bag contains one red disc and one blue disc. In a game of chance a
- player takes a disc at random and its colour is noted. After each turn the
- disc is returned to the bag, an extra red disc is added, and another disc
- is taken at random.
- The player pays £1 to play and wins if they have taken more blue discs
- than red discs at the end of the game.
- If the game is played for four turns, the probability of a player winning
- is exactly 11/120, and so the maximum prize fund the banker should
- allocate for winning in this game would be £10 before they would expect to
- incur a loss. Note that any payout will be a whole number of pounds and
- also includes the original £1 paid to play the game, so in the example
- given the player actually wins £9.
- Find the maximum prize fund that should be allocated to a single game in
- which fifteen turns are played.
- Answer: 2269
- Problem 122
- ===========
- The most naive way of computing n^15 requires fourteen multiplications:
- n × n × ... × n = n^15
- But using a "binary" method you can compute it in six multiplications:
- n × n = n^2
- n^2 × n^2 = n^4
- n^4 × n^4 = n^8
- n^8 × n^4 = n^12
- n^12 × n^2 = n^14
- n^14 × n = n^15
- However it is yet possible to compute it in only five multiplications:
- n × n = n^2
- n^2 × n = n^3
- n^3 × n^3 = n^6
- n^6 × n^6 = n^12
- n^12 × n^3 = n^15
- We shall define m(k) to be the minimum number of multiplications to
- compute n^k; for example m(15) = 5.
- For 1 ≤ k ≤ 200, find ∑ m(k).
- Answer: 1582
- Problem 123
- ===========
- Let p[n] be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder
- when (p[n]−1)^n + (p[n]+1)^n is divided by p[n]^2.
- For example, when n = 3, p[3] = 5, and 4^3 + 6^3 = 280 ≡ 5 mod 25.
- The least value of n for which the remainder first exceeds 10^9 is 7037.
- Find the least value of n for which the remainder first exceeds 10^10.
- Answer: 21035
- Problem 124
- ===========
- The radical of n, rad(n), is the product of the distinct prime factors of
- n. For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42.
- If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and
- sorting on n if the radical values are equal, we get:
- Unsorted Sorted
- n rad(n) n rad(n) k
- 1 1 1 1 1
- 2 2 2 2 2
- 3 3 4 2 3
- 4 2 8 2 4
- 5 5 3 3 5
- 6 6 9 3 6
- 7 7 5 5 7
- 8 2 6 6 8
- 9 3 7 7 9
- 10 10 10 10 10
- Let E(k) be the kth element in the sorted n column; for example, E(4) = 8
- and E(6) = 9.
- If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).
- Answer: 21417
- Problem 125
- ===========
- The palindromic number 595 is interesting because it can be written as the
- sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.
- There are exactly eleven palindromes below one-thousand that can be
- written as consecutive square sums, and the sum of these palindromes is
- 4164. Note that 1 = 0^2 + 1^2 has not been included as this problem is
- concerned with the squares of positive integers.
- Find the sum of all the numbers less than 10^8 that are both palindromic
- and can be written as the sum of consecutive squares.
- Answer: 1b5635e8ab723e01570ca783129493dd
- Problem 126
- ===========
- The minimum number of cubes to cover every visible face on a cuboid
- measuring 3 x 2 x 1 is twenty-two.
- If we then add a second layer to this solid it would require forty-six
- cubes to cover every visible face, the third layer would require
- seventy-eight cubes, and the fourth layer would require one-hundred and
- eighteen cubes to cover every visible face.
- However, the first layer on a cuboid measuring 5 x 1 x 1 also requires
- twenty-two cubes; similarly the first layer on cuboids measuring
- 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.
- We shall define C(n) to represent the number of cuboids that contain n
- cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118)
- = 8.
- It turns out that 154 is the least value of n for which C(n) = 10.
- Find the least value of n for which C(n) = 1000.
- p_126.gif
- Answer: 18522
- Problem 127
- ===========
- The radical of n, rad(n), is the product of distinct prime factors of n.
- For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42.
- We shall define the triplet of positive integers (a, b, c) to be an
- abc-hit if:
- 1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
- 2. a < b
- 3. a + b = c
- 4. rad(abc) < c
- For example, (5, 27, 32) is an abc-hit, because:
- 1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
- 2. 5 < 27
- 3. 5 + 27 = 32
- 4. rad(4320) = 30 < 32
- It turns out that abc-hits are quite rare and there are only thirty-one
- abc-hits for c < 1000, with ∑c = 12523.
- Find ∑c for c < 120000.
- Answer: 18407904
- Problem 128
- ===========
- A hexagonal tile with number 1 is surrounded by a ring of six hexagonal
- tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an
- anti-clockwise direction.
- New rings are added in the same fashion, with the next rings being
- numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows
- the first three rings.
- By finding the difference between tile n and each its six neighbours we
- shall define PD(n) to be the number of those differences which are prime.
- For example, working clockwise around tile 8 the differences are 12, 29,
- 11, 6, 1, and 13. So PD(8) = 3.
- In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and
- 10, hence PD(17) = 2.
- It can be shown that the maximum value of PD(n) is 3.
- If all of the tiles for which PD(n) = 3 are listed in ascending order to
- form a sequence, the 10th tile would be 271.
- Find the 2000th tile in this sequence.
- p_128.gif
- Answer: 93a1925da4792b4fa5d2dbb6ebb7c4a2
- Problem 129
- ===========
- A number consisting entirely of ones is called a repunit. We shall define
- R(k) to be a repunit of length k; for example, R(6) = 111111.
- Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
- that there always exists a value, k, for which R(k) is divisible by n, and
- let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
- 5.
- The least value of n for which A(n) first exceeds ten is 17.
- Find the least value of n for which A(n) first exceeds one-million.
- Answer: 1000023
- Problem 130
- ===========
- A number consisting entirely of ones is called a repunit. We shall define
- R(k) to be a repunit of length k; for example, R(6) = 111111.
- Given that n is a positive integer and GCD(n, 10) = 1, it can be shown
- that there always exists a value, k, for which R(k) is divisible by n, and
- let A(n) be the least such value of k; for example, A(7) = 6 and A(41) =
- 5.
- You are given that for all primes, p > 5, that p − 1 is divisible by A(p).
- For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
- However, there are rare composite values for which this is also true; the
- first five examples being 91, 259, 451, 481, and 703.
- Find the sum of the first twenty-five composite values of n for which
- GCD(n, 10) = 1 and n − 1 is divisible by A(n).
- Answer: 149253
- Problem 131
- ===========
- There are some prime values, p, for which there exists a positive integer,
- n, such that the expression n^3 + n^2p is a perfect cube.
- For example, when p = 19, 8^3 + 8^2×19 = 12^3.
- What is perhaps most surprising is that for each prime with this property
- the value of n is unique, and there are only four such primes below
- one-hundred.
- How many primes below one million have this remarkable property?
- Answer: 173
- Problem 132
- ===========
- A number consisting entirely of ones is called a repunit. We shall define
- R(k) to be a repunit of length k.
- For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these
- prime factors is 9414.
- Find the sum of the first forty prime factors of R(10^9).
- Answer: 843296
- Problem 133
- ===========
- A number consisting entirely of ones is called a repunit. We shall define
- R(k) to be a repunit of length k; for example, R(6) = 111111.
- Let us consider repunits of the form R(10^n).
- Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is
- divisible by 17. Yet there is no value of n for which R(10^n) will divide
- by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four
- primes below one-hundred that can be a factor of R(10^n).
- Find the sum of all the primes below one-hundred thousand that will never
- be a factor of R(10^n).
- Answer: c1d33d79d08cde65eaa78e4583ea0594
- Problem 134
- ===========
- Consider the consecutive primes p[1] = 19 and p[2] = 23. It can be
- verified that 1219 is the smallest number such that the last digits are
- formed by p[1] whilst also being divisible by p[2].
- In fact, with the exception of p[1] = 3 and p[2] = 5, for every pair of
- consecutive primes, p[2] > p[1], there exist values of n for which the
- last digits are formed by p[1] and n is divisible by p[2]. Let S be the
- smallest of these values of n.
- Find ∑ S for every pair of consecutive primes with 5 ≤ p[1] ≤ 1000000.
- Answer: f12b07460d2586ea47b4d305ae0b0539
- Problem 135
- ===========
- Given the positive integers, x, y, and z, are consecutive terms of an
- arithmetic progression, the least value of the positive integer, n, for
- which the equation, x^2 − y^2 − z^2 = n, has exactly two solutions is n =
- 27:
- 34^2 − 27^2 − 20^2 = 12^2 − 9^2 − 6^2 = 27
- It turns out that n = 1155 is the least value which has exactly ten
- solutions.
- How many values of n less than one million have exactly ten distinct
- solutions?
- Answer: 4989
- Problem 136
- ===========
- The positive integers, x, y, and z, are consecutive terms of an arithmetic
- progression. Given that n is a positive integer, the equation, x^2 − y^2 −
- z^2 = n, has exactly one solution when n = 20:
- 13^2 − 10^2 − 7^2 = 20
- In fact there are twenty-five values of n below one hundred for which the
- equation has a unique solution.
- How many values of n less than fifty million have exactly one solution?
- Answer: 2544559
- Problem 137
- ===========
- Consider the infinite polynomial series A[F](x) = xF[1] + x^2F[2] +
- x^3F[3] + ..., where F[k] is the kth term in the Fibonacci sequence: 1, 1,
- 2, 3, 5, 8, ... ; that is, F[k] = F[k−1] + F[k−2], F[1] = 1 and F[2] = 1.
- For this problem we shall be interested in values of x for which A[F](x)
- is a positive integer.
- Surprisingly A[F](1/2) = (1/2).1 + (1/2)^2.1 + (1/2)^3.2 + (1/2)^4.3 +
- (1/2)^5.5 + ...
- = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
- = 2
- The corresponding values of x for the first five natural numbers are shown
- below.
- ┌─────────┬───────┐
- │x │A[F](x)│
- ├─────────┼───────┤
- │√2−1 │1 │
- ├─────────┼───────┤
- │1/2 │2 │
- ├─────────┼───────┤
- │(√13−2)/3│3 │
- ├─────────┼───────┤
- │(√89−5)/8│4 │
- ├─────────┼───────┤
- │(√34−3)/5│5 │
- └─────────┴───────┘
- We shall call A[F](x) a golden nugget if x is rational, because they
- become increasingly rarer; for example, the 10th golden nugget is
- 74049690.
- Find the 15th golden nugget.
- Answer: 44845aa0f47ec925a3b43e6460a55e27
- Problem 138
- ===========
- Consider the isosceles triangle with base length, b = 16, and legs, L =
- 17.
- By using the Pythagorean theorem it can be seen that the height of the
- triangle, h = √(17^2 − 8^2) = 15, which is one less than the base length.
- With b = 272 and L = 305, we get h = 273, which is one more than the base
- length, and this is the second smallest isosceles triangle with the
- property that h = b ± 1.
- Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1
- and b, L are positive integers.
- p_138.gif
- Answer: f7524f4d0d6d042c0f92a0d6469aff85
- Problem 139
- ===========
- Let (a, b, c) represent the three sides of a right angle triangle with
- integral length sides. It is possible to place four such triangles
- together to form a square with length c.
- For example, (3, 4, 5) triangles can be placed together to form a 5 by 5
- square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5
- square can be tiled with twenty-five 1 by 1 squares.
- However, if (5, 12, 13) triangles were used then the hole would measure 7
- by 7 and these could not be used to tile the 13 by 13 square.
- Given that the perimeter of the right triangle is less than one-hundred
- million, how many Pythagorean triangles would allow such a tiling to take
- place?
- p_139.gif
- Answer: 10057761
- Problem 140
- ===========
- Consider the infinite polynomial series A[G](x) = xG[1] + x^2G[2] +
- x^3G[3] + ..., where G[k] is the kth term of the second order recurrence
- relation G[k] = G[k−1] + G[k−2], G[1] = 1 and G[2] = 4; that is, 1, 4, 5,
- 9, 14, 23, ... .
- For this problem we shall be concerned with values of x for which A[G](x)
- is a positive integer.
- The corresponding values of x for the first five natural numbers are shown
- below.
- ┌───────────┬───────┐
- │x │A[G](x)│
- ├───────────┼───────┤
- │(√5−1)/4 │1 │
- ├───────────┼───────┤
- │2/5 │2 │
- ├───────────┼───────┤
- │(√22−2)/6 │3 │
- ├───────────┼───────┤
- │(√137−5)/14│4 │
- ├───────────┼───────┤
- │1/2 │5 │
- └───────────┴───────┘
- We shall call A[G](x) a golden nugget if x is rational, because they
- become increasingly rarer; for example, the 20th golden nugget is
- 211345365.
- Find the sum of the first thirty golden nuggets.
- Answer: e5d75f96929ba250b2732aad52f3028c
- Problem 141
- ===========
- A positive integer, n, is divided by d and the quotient and remainder are
- q and r respectively. In addition d, q, and r are consecutive positive
- integer terms in a geometric sequence, but not necessarily in that order.
- For example, 58 divided by 6 has quotient 9 and remainder 4. It can also
- be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common
- ratio 3/2).
- We will call such numbers, n, progressive.
- Some progressive numbers, such as 9 and 10404 = 102^2, happen to also be
- perfect squares.
- The sum of all progressive perfect squares below one hundred thousand is
- 124657.
- Find the sum of all progressive perfect squares below one trillion
- (10^12).
- Answer: 2aaefa1db80951be140183f9e8c0194e
- Problem 142
- ===========
- Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x
- − y, x + z, x − z, y + z, y − z are all perfect squares.
- Answer: 1006193
- Problem 143
- ===========
- Let ABC be a triangle with all interior angles being less than 120
- degrees. Let X be any point inside the triangle and let XA = p, XC = q,
- and XB = r.
- Fermat challenged Torricelli to find the position of X such that p + q + r
- was minimised.
- Torricelli was able to prove that if equilateral triangles AOB, BNC and
- AMC are constructed on each side of triangle ABC, the circumscribed
- circles of AOB, BNC, and AMC will intersect at a single point, T, inside
- the triangle. Moreover he proved that T, called the Torricelli/Fermat
- point, minimises p + q + r. Even more remarkable, it can be shown that
- when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO
- also intersect at T.
- If the sum is minimised and a, b, c, p, q and r are all positive integers
- we shall call triangle ABC a Torricelli triangle. For example, a = 399, b
- = 455, c = 511 is an example of a Torricelli triangle, with p + q + r =
- 784.
- Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli
- triangles.
- p_143_torricelli.gif
- Answer: 30758397
- Problem 144
- ===========
- In laser physics, a "white cell" is a mirror system that acts as a delay
- line for the laser beam. The beam enters the cell, bounces around on the
- mirrors, and eventually works its way back out.
- The specific white cell we will be considering is an ellipse with the
- equation 4x^2 + y^2 = 100
- The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing,
- allowing the light to enter and exit through the hole.
- The light beam in this problem starts at the point (0.0,10.1) just outside
- the white cell, and the beam first impacts the mirror at (1.4,-9.6).
- Each time the laser beam hits the surface of the ellipse, it follows the
- usual law of reflection "angle of incidence equals angle of reflection."
- That is, both the incident and reflected beams make the same angle with
- the normal line at the point of incidence.
- In the figure on the left, the red line shows the first two points of
- contact between the laser beam and the wall of the white cell; the blue
- line shows the line tangent to the ellipse at the point of incidence of
- the first bounce.
- The slope m of the tangent line at any point (x,y) of the given ellipse
- is: m = −4x/y
- The normal line is perpendicular to this tangent line at the point of
- incidence.
- The animation on the right shows the first 10 reflections of the beam.
- How many times does the beam hit the internal surface of the white cell
- before exiting?
- p_144_1.gif
- p_144_2.gif
- Answer: 354
- Problem 145
- ===========
- Some positive integers n have the property that the sum [ n + reverse(n) ]
- consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and
- 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409,
- and 904 are reversible. Leading zeroes are not allowed in either n or
- reverse(n).
- There are 120 reversible numbers below one-thousand.
- How many reversible numbers are there below one-billion (10^9)?
- Answer: 608720
- Problem 146
- ===========
- The smallest positive integer n for which the numbers n^2+1, n^2+3, n^2+7,
- n^2+9, n^2+13, and n^2+27 are consecutive primes is 10. The sum of all
- such integers n below one-million is 1242490.
- What is the sum of all such integers n below 150 million?
- Answer: 525bd2bf0e31b0f19b38a1d21f2f6a16
- Problem 147
- ===========
- In a 3x2 cross-hatched grid, a total of 37 different rectangles could be
- situated within that grid as indicated in the sketch.
- There are 5 grids smaller than 3x2, vertical and horizontal dimensions
- being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is
- cross-hatched, the following number of different rectangles could be
- situated within those smaller grids:
- 1x1: 1
- 2x1: 4
- 3x1: 8
- 1x2: 4
- 2x2: 18
- Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles
- could be situated within 3x2 and smaller grids.
- How many different rectangles could be situated within 47x43 and smaller
- grids?
- p_147.gif
- Answer: d0fca7d85d4a4df043a2ae5772ea472e
- Problem 148
- ===========
- We can easily verify that none of the entries in the first seven rows of
- Pascal's triangle are divisible by 7:
- 1
- 1 1
- 1 2 1
- 1 3 3 1
- 1 4 6 4 1
- 1 5 10 10 5 1
- 1 6 15 20 15 6 1
- However, if we check the first one hundred rows, we will find that only
- 2361 of the 5050 entries are not divisible by 7.
- Find the number of entries which are not divisible by 7 in the first one
- billion (10^9) rows of Pascal's triangle.
- Answer: 8a631ab4e3d06baf88299bf4e501b837
- Problem 149
- ===========
- Looking at the table below, it is easy to verify that the maximum possible
- sum of adjacent numbers in any direction (horizontal, vertical, diagonal
- or anti-diagonal) is 16 (= 8 + 7 + 1).
- ┌────┬────┬────┬─────┐
- │ −2 │ 5 │ 3 │ 2 │
- ├────┼────┼────┼─────┤
- │ 9 │ −6 │ 5 │ 1 │
- ├────┼────┼────┼─────┤
- │ 3 │ 2 │ 7 │ 3 │
- ├────┼────┼────┼─────┤
- │ −1 │ 8 │ −4 │ 8 │
- └────┴────┴────┴─────┘
- Now, let us repeat the search, but on a much larger scale:
- First, generate four million pseudo-random numbers using a specific form
- of what is known as a "Lagged Fibonacci Generator":
- For 1 ≤ k ≤ 55, s[k] = [100003 − 200003k + 300007k^3] (modulo 1000000) −
- 500000.
- For 56 ≤ k ≤ 4000000, s[k] = [s[k−24] + s[k−55] + 1000000] (modulo
- 1000000) − 500000.
- Thus, s[10] = −393027 and s[100] = 86613.
- The terms of s are then arranged in a 2000×2000 table, using the first
- 2000 numbers to fill the first row (sequentially), the next 2000 numbers
- to fill the second row, and so on.
- Finally, find the greatest sum of (any number of) adjacent entries in any
- direction (horizontal, vertical, diagonal or anti-diagonal).
- Answer: 52852124
- Problem 150
- ===========
- In a triangular array of positive and negative integers, we wish to find a
- sub-triangle such that the sum of the numbers it contains is the smallest
- possible.
- In the example below, it can be easily verified that the marked triangle
- satisfies this condition having a sum of −42.
- We wish to make such a triangular array with one thousand rows, so we
- generate 500500 pseudo-random numbers s[k] in the range ±2^19, using a
- type of random number generator (known as a Linear Congruential Generator)
- as follows:
- t := 0
- for k = 1 up to k = 500500:
- t := (615949*t + 797807) modulo 2^20
- s[k] := t−2^19
- Thus: s[1] = 273519, s[2] = −153582, s[3] = 450905 etc
- Our triangular array is then formed using the pseudo-random numbers thus:
- s[1]
- s[2] s[3]
- s[4] s[5] s[6]
- s[7] s[8] s[9] s[10]
- ...
- Sub-triangles can start at any element of the array and extend down as far
- as we like (taking-in the two elements directly below it from the next
- row, the three elements directly below from the row after that, and so
- on).
- The "sum of a sub-triangle" is defined as the sum of all the elements it
- contains.
- Find the smallest possible sub-triangle sum.
- p_150.gif
- Answer: 1802939e514020769701c59b422c0498
- Problem 151
- ===========
- A printing shop runs 16 batches (jobs) every week and each batch requires
- a sheet of special colour-proofing paper of size A5.
- Every Monday morning, the foreman opens a new envelope, containing a large
- sheet of the special paper with size A1.
- He proceeds to cut it in half, thus getting two sheets of size A2. Then he
- cuts one of them in half to get two sheets of size A3 and so on until he
- obtains the A5-size sheet needed for the first batch of the week.
- All the unused sheets are placed back in the envelope.
- At the beginning of each subsequent batch, he takes from the envelope one
- sheet of paper at random. If it is of size A5, he uses it. If it is
- larger, he repeats the 'cut-in-half' procedure until he has what he needs
- and any remaining sheets are always placed back in the envelope.
- Excluding the first and last batch of the week, find the expected number
- of times (during each week) that the foreman finds a single sheet of paper
- in the envelope.
- Give your answer rounded to six decimal places using the format x.xxxxxx .
- p_151.gif
- Answer: 0.464399
- Problem 152
- ===========
- There are several ways to write the number 1/2 as a sum of inverse squares
- using distinct integers.
- For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:
- In fact, only using integers between 2 and 45 inclusive, there are exactly
- three ways to do it, the remaining two being:
- {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}.
- How many ways are there to write the number 1/2 as a sum of inverse
- squares using distinct integers between 2 and 80 inclusive?
- p_152_sum.gif
- Answer: 301
- Problem 153
- ===========
- As we all know the equation x^2=-1 has no solutions for real x.
- If we however introduce the imaginary number i this equation has two
- solutions: x=i and x=-i.
- If we go a step further the equation (x-3)^2=-4 has two complex solutions:
- x=3+2i and x=3-2i.
- x=3+2i and x=3-2i are called each others' complex conjugate.
- Numbers of the form a+bi are called complex numbers.
- In general a+bi and a−bi are each other's complex conjugate.
- A Gaussian Integer is a complex number a+bi such that both a and b are
- integers.
- The regular integers are also Gaussian integers (with b=0).
- To distinguish them from Gaussian integers with b ≠ 0 we call such
- integers "rational integers."
- A Gaussian integer is called a divisor of a rational integer n if the
- result is also a Gaussian integer.
- If for example we divide 5 by 1+2i we can simplify in the following
- manner:
- Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
- The result is .
- So 1+2i is a divisor of 5.
- Note that 1+i is not a divisor of 5 because .
- Note also that if the Gaussian Integer (a+bi) is a divisor of a rational
- integer n, then its complex conjugate (a−bi) is also a divisor of n.
- In fact, 5 has six divisors such that the real part is positive: {1, 1 +
- 2i, 1 − 2i, 2 + i, 2 − i, 5}.
- The following is a table of all of the divisors for the first five
- positive rational integers:
- ┌───┬──────────────────────────────┬───────────────┐
- │ n │ Gaussian integer divisors │ Sum s(n) of │
- │ │ with positive real part │ thesedivisors │
- ├───┼──────────────────────────────┼───────────────┤
- │ 1 │ 1 │ 1 │
- ├───┼──────────────────────────────┼───────────────┤
- │ 2 │ 1, 1+i, 1-i, 2 │ 5 │
- ├───┼──────────────────────────────┼───────────────┤
- │ 3 │ 1, 3 │ 4 │
- ├───┼──────────────────────────────┼───────────────┤
- │ 4 │ 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 │ 13 │
- ├───┼──────────────────────────────┼───────────────┤
- │ 5 │ 1, 1+2i, 1-2i, 2+i, 2-i, 5 │ 12 │
- └───┴──────────────────────────────┴───────────────┘
- For divisors with positive real parts, then, we have: .
- For 1 ≤ n ≤ 10^5, ∑ s(n)=17924657155.
- What is ∑ s(n) for 1 ≤ n ≤ 10^8?
- p_153_formule1.gif
- p_153_formule2.gif
- p_153_formule5.gif
- p_153_formule6.gif
- Answer: 08ec9d6e6c2275d37e7a227fb2d1f06f
- Problem 154
- ===========
- A triangular pyramid is constructed using spherical balls so that each
- ball rests on exactly three balls of the next lower level.
- Then, we calculate the number of paths leading from the apex to each
- position:
- A path starts at the apex and progresses downwards to any of the three
- spheres directly below the current position.
- Consequently, the number of paths to reach a certain position is the sum
- of the numbers immediately above it (depending on the position, there are
- up to three numbers above it).
- The result is Pascal's pyramid and the numbers at each level n are the
- coefficients of the trinomial expansion (x + y + z)^n.
- How many coefficients in the expansion of (x + y + z)^200000 are multiples
- of 10^12?
- p_154_pyramid.gif
- Answer: de866633fa075beb3897cbbc8abf2400
- Problem 155
- ===========
- An electric circuit uses exclusively identical capacitors of the same
- value C.
- The capacitors can be connected in series or in parallel to form
- sub-units, which can then be connected in series or in parallel with other
- capacitors or other sub-units to form larger sub-units, and so on up to a
- final circuit.
- Using this simple procedure and up to n identical capacitors, we can make
- circuits having a range of different total capacitances. For example,
- using up to n=3 capacitors of 60 F each, we can obtain the following 7
- distinct total capacitance values:
- If we denote by D(n) the number of distinct total capacitance values we
- can obtain when using up to n equal-valued capacitors and the simple
- procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ...
- Find D(18).
- Reminder : When connecting capacitors C[1], C[2] etc in parallel, the
- total capacitance is C[T] = C[1] + C[2] +...,
- whereas when connecting them in series, the overall capacitance is given
- by:
- p_155_capsmu.gif
- p_155_capacitors1.gif
- p_155_capsform.gif
- Answer: 3857447
- Problem 156
- ===========
- Starting from zero the natural numbers are written down in base 10 like
- this:
- 0 1 2 3 4 5 6 7 8 9 10 11 12....
- Consider the digit d=1. After we write down each number n, we will update
- the number of ones that have occurred and call this number f(n,1). The
- first values for f(n,1), then, are as follows:
- n f(n,1)
- 0 0
- 1 1
- 2 1
- 3 1
- 4 1
- 5 1
- 6 1
- 7 1
- 8 1
- 9 1
- 10 2
- 11 4
- 12 5
- Note that f(n,1) never equals 3.
- So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The
- next solution is n=199981.
- In the same manner the function f(n,d) gives the total number of digits d
- that have been written down after the number n has been written.
- In fact, for every digit d ≠ 0, 0 is the first solution of the equation
- f(n,d)=n.
- Let s(d) be the sum of all the solutions for which f(n,d)=n.
- You are given that s(1)=22786974071.
- Find ∑ s(d) for 1 ≤ d ≤ 9.
- Note: if, for some n, f(n,d)=n for more than one value of d this value of
- n is counted again for every value of d for which f(n,d)=n.
- Answer: ac0c6b67ed28cebb02b802e7a204aaee
- Problem 157
- ===========
- Consider the diophantine equation ^1/[a]+^1/[b]= ^p/[10^n] with a, b, p, n
- positive integers and a ≤ b.
- For n=1 this equation has 20 solutions that are listed below:
- 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
- 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
- 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
- 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
- How many solutions has this equation for 1 ≤ n ≤ 9?
- Answer: 53490
- Problem 158
- ===========
- Taking three different letters from the 26 letters of the alphabet,
- character strings of length three can be formed.
- Examples are 'abc', 'hat' and 'zyx'.
- When we study these three examples we see that for 'abc' two characters
- come lexicographically after its neighbour to the left.
- For 'hat' there is exactly one character that comes lexicographically
- after its neighbour to the left. For 'zyx' there are zero characters that
- come lexicographically after its neighbour to the left.
- In all there are 10400 strings of length 3 for which exactly one character
- comes lexicographically after its neighbour to the left.
- We now consider strings of n ≤ 26 different characters from the alphabet.
- For every n, p(n) is the number of strings of length n for which exactly
- one character comes lexicographically after its neighbour to the left.
- What is the maximum value of p(n)?
- Answer: 6070fa194890e52b2989af5b542aee90
- Problem 159
- ===========
- A composite number can be factored many different ways. For instance, not
- including multiplication by one, 24 can be factored in 7 distinct ways:
- 24 = 2x2x2x3
- 24 = 2x3x4
- 24 = 2x2x6
- 24 = 4x6
- 24 = 3x8
- 24 = 2x12
- 24 = 24
- Recall that the digital root of a number, in base 10, is found by adding
- together the digits of that number, and repeating that process until a
- number is arrived at that is less than 10. Thus the digital root of 467 is
- 8.
- We shall call a Digital Root Sum (DRS) the sum of the digital roots of the
- individual factors of our number.
- The chart below demonstrates all of the DRS values for 24.
- ┌─────────────┬────────────────┐
- │Factorisation│Digital Root Sum│
- ├─────────────┼────────────────┤
- │2x2x2x3 │ 9 │
- ├─────────────┼────────────────┤
- │2x3x4 │ 9 │
- ├─────────────┼────────────────┤
- │2x2x6 │ 10 │
- ├─────────────┼────────────────┤
- │4x6 │ 10 │
- ├─────────────┼────────────────┤
- │3x8 │ 11 │
- ├─────────────┼────────────────┤
- │2x12 │ 5 │
- ├─────────────┼────────────────┤
- │24 │ 6 │
- └─────────────┴────────────────┘
- The maximum Digital Root Sum of 24 is 11.
- The function mdrs(n) gives the maximum Digital Root Sum of n. So
- mdrs(24)=11.
- Find ∑mdrs(n) for 1 < n < 1,000,000.
- Answer: 14489159
- Problem 160
- ===========
- For any N, let f(N) be the last five digits before the trailing zeroes in
- N!.
- For example,
- 9! = 362880 so f(9)=36288
- 10! = 3628800 so f(10)=36288
- 20! = 2432902008176640000 so f(20)=17664
- Find f(1,000,000,000,000)
- Answer: 16576
- Problem 161
- ===========
- A triomino is a shape consisting of three squares joined via the
- edges.There are two basic forms:
- If all possible orientations are taken into account there are six:
- Any n by m grid for which nxm is divisible by 3 can be tiled with
- triominoes.
- If we consider tilings that can be obtained by reflection or rotation from
- another tiling as different there are 41 ways a 2 by 9 grid can be tiled
- with triominoes:
- In how many ways can a 9 by 12 grid be tiled in this way by triominoes?
- p_161_trio1.gif
- p_161_trio3.gif
- p_161_k9.gif
- Answer: 975ccc38bb5402c5b485f3de5928d919
- Problem 162
- ===========
- In the hexadecimal number system numbers are represented using 16
- different digits:
- 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
- The hexadecimal number AF when written in the decimal number system equals
- 10x16+15=175.
- In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1
- and A are all present.
- Like numbers written in base ten we write hexadecimal numbers without
- leading zeroes.
- How many hexadecimal numbers containing at most sixteen hexadecimal digits
- exist with all of the digits 0,1, and A present at least once?
- Give your answer as a hexadecimal number.
- (A,B,C,D,E and F in upper case, without any leading or trailing code that
- marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and
- not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F)
- Answer: 049419b9fdad9af74d5888626fff56a3
- Problem 163
- ===========
- Consider an equilateral triangle in which straight lines are drawn from
- each vertex to the middle of the opposite side, such as in the size 1
- triangle in the sketch below.
- Sixteen triangles of either different shape or size or orientation or
- location can now be observed in that triangle. Using size 1 triangles as
- building blocks, larger triangles can be formed, such as the size 2
- triangle in the above sketch. One-hundred and four triangles of either
- different shape or size or orientation or location can now be observed in
- that size 2 triangle.
- It can be observed that the size 2 triangle contains 4 size 1 triangle
- building blocks. A size 3 triangle would contain 9 size 1 triangle
- building blocks and a size n triangle would thus contain n^2 size 1
- triangle building blocks.
- If we denote T(n) as the number of triangles present in a triangle of size
- n, then
- T(1) = 16
- T(2) = 104
- Find T(36).
- p_163.gif
- Answer: 343047
- Problem 164
- ===========
- How many 20 digit numbers n (without any leading zero) exist such that no
- three consecutive digits of n have a sum greater than 9?
- Answer: 6e96debf3bfe7cc132401bafe5a5d6d6
- Problem 165
- ===========
- A segment is uniquely defined by its two endpoints.
- By considering two line segments in plane geometry there are three
- possibilities:
- the segments have zero points, one point, or infinitely many points in
- common.
- Moreover when two segments have exactly one point in common it might be
- the case that that common point is an endpoint of either one of the
- segments or of both. If a common point of two segments is not an endpoint
- of either of the segments it is an interior point of both segments.
- We will call a common point T of two segments L[1] and L[2] a true
- intersection point of L[1] and L[2] if T is the only common point of L[1]
- and L[2] and T is an interior point of both segments.
- Consider the three segments L[1], L[2], and L[3]:
- L[1]: (27, 44) to (12, 32)
- L[2]: (46, 53) to (17, 62)
- L[3]: (46, 70) to (22, 40)
- It can be verified that line segments L[2] and L[3] have a true
- intersection point. We note that as the one of the end points of L[3]:
- (22,40) lies on L[1] this is not considered to be a true point of
- intersection. L[1] and L[2] have no common point. So among the three line
- segments, we find one true intersection point.
- Now let us do the same for 5000 line segments. To this end, we generate
- 20000 numbers using the so-called "Blum Blum Shub" pseudo-random number
- generator.
- s[0] = 290797
- s[n+1] = s[n]×s[n] (modulo 50515093)
- t[n] = s[n] (modulo 500)
- To create each line segment, we use four consecutive numbers t[n]. That
- is, the first line segment is given by:
- (t[1], t[2]) to (t[3], t[4])
- The first four numbers computed according to the above generator should
- be: 27, 144, 12 and 232. The first segment would thus be (27,144) to
- (12,232).
- How many distinct true intersection points are found among the 5000 line
- segments?
- Answer: 2868868
- Problem 166
- ===========
- A 4x4 grid is filled with digits d, 0 ≤ d ≤ 9.
- It can be seen that in the grid
- 6 3 3 0
- 5 0 4 3
- 0 7 1 4
- 1 2 4 5
- the sum of each row and each column has the value 12. Moreover the sum of
- each diagonal is also 12.
- In how many ways can you fill a 4x4 grid with the digits d, 0 ≤ d ≤ 9 so
- that each row, each column, and both diagonals have the same sum?
- Answer: 7130034
- Problem 167
- ===========
- For two positive integers a and b, the Ulam sequence U(a,b) is defined by
- U(a,b)[1] = a, U(a,b)[2] = b and for k > 2,U(a,b)[k] is the smallest
- integer greater than U(a,b)[(k-1)] which can be written in exactly one way
- as the sum of two distinct previous members of U(a,b).
- For example, the sequence U(1,2) begins with
- 1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;
- 5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations
- as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.
- Find ∑U(2,2n+1)[k] for 2 ≤ n ≤10, where k = 10^11.
- Answer: aa5b61f6f4d96cbaeb5944b8fcdf64a3
- Problem 168
- ===========
- Consider the number 142857. We can right-rotate this number by moving the
- last digit (7) to the front of it, giving us 714285.
- It can be verified that 714285=5×142857.
- This demonstrates an unusual property of 142857: it is a divisor of its
- right-rotation.
- Find the last 5 digits of the sum of all integers n, 10 < n < 10^100, that
- have this property.
- Answer: 59206
- Problem 169
- ===========
- Define f(0)=1 and f(n) to be the number of different ways n can be
- expressed as a sum of integer powers of 2 using each power no more than
- twice.
- For example, f(10)=5 since there are five different ways to express 10:
- 1 + 1 + 8
- 1 + 1 + 4 + 4
- 1 + 1 + 2 + 2 + 4
- 2 + 4 + 4
- 2 + 8
- What is f(10^25)?
- Answer: d149d4836703a8908becea56ddd3ed42
- Problem 170
- ===========
- Take the number 6 and multiply it by each of 1273 and 9854:
- 6 × 1273 = 7638
- 6 × 9854 = 59124
- By concatenating these products we get the 1 to 9 pandigital 763859124. We
- will call 763859124 the "concatenated product of 6 and (1273,9854)".
- Notice too, that the concatenation of the input numbers, 612739854, is
- also 1 to 9 pandigital.
- The same can be done for 0 to 9 pandigital numbers.
- What is the largest 0 to 9 pandigital 10-digit concatenated product of an
- integer with two or more other integers, such that the concatenation of
- the input numbers is also a 0 to 9 pandigital 10-digit number?
- Answer: 6ffe65352f717c1731666a107ace96c1
- Problem 171
- ===========
- For a positive integer n, let f(n) be the sum of the squares of the digits
- (in base 10) of n, e.g.
- f(3) = 3^2 = 9,
- f(25) = 2^2 + 5^2 = 4 + 25 = 29,
- f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36
- Find the last nine digits of the sum of all n, 0 < n < 10^20, such that
- f(n) is a perfect square.
- Answer: 142989277
- Problem 172
- ===========
- How many 18-digit numbers n (without leading zeros) are there such that no
- digit occurs more than three times in n?
- Answer: f5f260ee21ead7478403c2ccd18a1829
- Problem 173
- ===========
- We shall define a square lamina to be a square outline with a square
- "hole" so that the shape possesses vertical and horizontal symmetry. For
- example, using exactly thirty-two square tiles we can form two different
- square laminae:
- With one-hundred tiles, and not necessarily using all of the tiles at one
- time, it is possible to form forty-one different square laminae.
- Using up to one million tiles how many different square laminae can be
- formed?
- p_173_square_laminas.gif
- Answer: 1572729
- Problem 174
- ===========
- We shall define a square lamina to be a square outline with a square
- "hole" so that the shape possesses vertical and horizontal symmetry.
- Given eight tiles it is possible to form a lamina in only one way: 3x3
- square with a 1x1 hole in the middle. However, using thirty-two tiles it
- is possible to form two distinct laminae.
- If t represents the number of tiles used, we shall say that t = 8 is type
- L(1) and t = 32 is type L(2).
- Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for
- example, N(15) = 832.
- What is ∑ N(n) for 1 ≤ n ≤ 10?
- p_173_square_laminas.gif
- Answer: 209566
- Problem 175
- ===========
- Define f(0)=1 and f(n) to be the number of ways to write n as a sum of
- powers of 2 where no power occurs more than twice.
- For example, f(10)=5 since there are five different ways to express 10:
- 10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1
- It can be shown that for every fraction p/q (p>0, q>0) there exists at
- least one integer n such that
- f(n)/f(n-1)=p/q.
- For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241.
- The binary expansion of 241 is 11110001.
- Reading this binary number from the most significant bit to the least
- significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the
- string 4,3,1 the Shortened Binary Expansion of 241.
- Find the Shortened Binary Expansion of the smallest n for which
- f(n)/f(n-1)=123456789/987654321.
- Give your answer as comma separated integers, without any whitespaces.
- Answer: 796dddd004c3465229058072f5b4583e
- Problem 176
- ===========
- The four right-angled triangles with sides (9,12,15), (12,16,20),
- (5,12,13) and (12,35,37) all have one of the shorter sides (catheti) equal
- to 12. It can be shown that no other integer sided right-angled triangle
- exists with one of the catheti equal to 12.
- Find the smallest integer that can be the length of a cathetus of exactly
- 47547 different integer sided right-angled triangles.
- Answer: c47c782ebaf8cdbb60eebfa86cd0003c
- Problem 177
- ===========
- Let ABCD be a convex quadrilateral, with diagonals AC and BD. At each
- vertex the diagonal makes an angle with each of the two sides, creating
- eight corner angles.
- For example, at vertex A, the two angles are CAD, CAB.
- We call such a quadrilateral for which all eight corner angles have
- integer values when measured in degrees an "integer angled quadrilateral".
- An example of an integer angled quadrilateral is a square, where all eight
- corner angles are 45°. Another example is given by DAC = 20°, BAC = 60°,
- ABD = 50°, CBD = 30°, BCA = 40°, DCA = 30°, CDB = 80°, ADB = 50°.
- What is the total number of non-similar integer angled quadrilaterals?
- Note: In your calculations you may assume that a calculated angle is
- integral if it is within a tolerance of 10^-9 of an integer value.
- p_177_quad.gif
- Answer: 129325
- Problem 178
- ===========
- Consider the number 45656.
- It can be seen that each pair of consecutive digits of 45656 has a
- difference of one.
- A number for which every pair of consecutive digits has a difference of
- one is called a step number.
- A pandigital number contains every decimal digit from 0 to 9 at least
- once.
- How many pandigital step numbers less than 10^40 are there?
- Answer: 2ffddfa898fa5df6321aebea84d4f33f
- Problem 179
- ===========
- Find the number of integers 1 < n < 10^7, for which n and n + 1 have the
- same number of positive divisors. For example, 14 has the positive
- divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.
- Answer: 986262
- Problem 180
- ===========
- For any integer n, consider the three functions
- f[1,n](x,y,z) = x^n+1 + y^n+1 − z^n+1
- f[2,n](x,y,z) = (xy + yz + zx)*(x^n-1 + y^n-1 − z^n-1)
- f[3,n](x,y,z) = xyz*(x^n-2 + y^n-2 − z^n-2)
- and their combination
- f[n](x,y,z) = f[1,n](x,y,z) + f[2,n](x,y,z) − f[3,n](x,y,z)
- We call (x,y,z) a golden triple of order k if x, y, and z are all rational
- numbers of the form a / b with
- 0 < a < b ≤ k and there is (at least) one integer n, so that f[n](x,y,z) =
- 0.
- Let s(x,y,z) = x + y + z.
- Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples
- (x,y,z) of order 35.
- All the s(x,y,z) and t must be in reduced form.
- Find u + v.
- Answer: 6459f69d151314c59df404868f45fa96
- Problem 181
- ===========
- Having three black objects B and one white object W they can be grouped in
- 7 ways like this:
- (BBBW) (B,BBW) (B,B,BW) (B,B,B,W) (B,BB,W) (BBB,W) (BB,BW)
- In how many ways can sixty black objects B and forty white objects W be
- thus grouped?
- Answer: 0e1233ecbc058dabf54a8602eac55d95
- Problem 182
- ===========
- The RSA encryption is based on the following procedure:
- Generate two distinct primes p and q.
- Compute n=pq and φ=(p-1)(q-1).
- Find an integer e, 1<e<φ, such that gcd(e,φ)=1.
- A message in this system is a number in the interval [0,n-1].
- A text to be encrypted is then somehow converted to messages (numbers in
- the interval [0,n-1]).
- To encrypt the text, for each message, m, c=m^e mod n is calculated.
- To decrypt the text, the following procedure is needed: calculate d such
- that ed=1 mod φ, then for each encrypted message, c, calculate m=c^d mod
- n.
- There exist values of e and m such that m^e mod n=m.
- We call messages m for which m^e mod n=m unconcealed messages.
- An issue when choosing e is that there should not be too many unconcealed
- messages.
- For instance, let p=19 and q=37.
- Then n=19*37=703 and φ=18*36=648.
- If we choose e=181, then, although gcd(181,648)=1 it turns out that all
- possible messages
- m (0≤m≤n-1) are unconcealed when calculating m^e mod n.
- For any valid choice of e there exist some unconcealed messages.
- It's important that the number of unconcealed messages is at a minimum.
- Choose p=1009 and q=3643.
- Find the sum of all values of e, 1<e<φ(1009,3643) and gcd(e,φ)=1, so that
- the number of unconcealed messages for this value of e is at a minimum.
- Answer: 088ad9a61e60b9309e91cfc3ed27d729
- Problem 183
- ===========
- Let N be a positive integer and let N be split into k equal parts, r =
- N/k, so that N = r + r + ... + r.
- Let P be the product of these parts, P = r × r × ... × r = r^k.
- For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 +
- 2.2 + 2.2, then P = 2.2^5 = 51.53632.
- Let M(N) = P[max] for a given value of N.
- It turns out that the maximum for N = 11 is found by splitting eleven into
- four equal parts which leads to P[max] = (11/4)^4; that is, M(11) =
- 14641/256 = 57.19140625, which is a terminating decimal.
- However, for N = 8 the maximum is achieved by splitting it into three
- equal parts, so M(8) = 512/27, which is a non-terminating decimal.
- Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is
- a terminating decimal.
- For example, ΣD(N) for 5 ≤ N ≤ 100 is 2438.
- Find ΣD(N) for 5 ≤ N ≤ 10000.
- Answer: 48861552
- Problem 184
- ===========
- Consider the set I[r] of points (x,y) with integer co-ordinates in the
- interior of the circle with radius r, centered at the origin, i.e. x^2 +
- y^2 < r^2.
- For a radius of 2, I[2] contains the nine points (0,0), (1,0), (1,1),
- (0,1), (-1,1), (-1,0), (-1,-1), (0,-1) and (1,-1). There are eight
- triangles having all three vertices in I[2] which contain the origin in
- the interior. Two of them are shown below, the others are obtained from
- these by rotation.
- For a radius of 3, there are 360 triangles containing the origin in the
- interior and having all vertices in I[3] and for I[5] the number is 10600.
- How many triangles are there containing the origin in the interior and
- having all three vertices in I[105]?
- p_184.gif
- Answer: aa80f8619ed594e5d7852564457dbca6
- Problem 185
- ===========
- The game Number Mind is a variant of the well known game Master Mind.
- Instead of coloured pegs, you have to guess a secret sequence of digits.
- After each guess you're only told in how many places you've guessed the
- correct digit. So, if the sequence was 1234 and you guessed 2036, you'd be
- told that you have one correct digit; however, you would NOT be told that
- you also have another digit in the wrong place.
- For instance, given the following guesses for a 5-digit secret sequence,
- 90342 ;2 correct
- 70794 ;0 correct
- 39458 ;2 correct
- 34109 ;1 correct
- 51545 ;2 correct
- 12531 ;1 correct
- The correct sequence 39542 is unique.
- Based on the following guesses,
- 5616185650518293 ;2 correct
- 3847439647293047 ;1 correct
- 5855462940810587 ;3 correct
- 9742855507068353 ;3 correct
- 4296849643607543 ;3 correct
- 3174248439465858 ;1 correct
- 4513559094146117 ;2 correct
- 7890971548908067 ;3 correct
- 8157356344118483 ;1 correct
- 2615250744386899 ;2 correct
- 8690095851526254 ;3 correct
- 6375711915077050 ;1 correct
- 6913859173121360 ;1 correct
- 6442889055042768 ;2 correct
- 2321386104303845 ;0 correct
- 2326509471271448 ;2 correct
- 5251583379644322 ;2 correct
- 1748270476758276 ;3 correct
- 4895722652190306 ;1 correct
- 3041631117224635 ;3 correct
- 1841236454324589 ;3 correct
- 2659862637316867 ;2 correct
- Find the unique 16-digit secret sequence.
- Answer: 70f84864f21c4bf07ee53436580cd4bb
- Problem 186
- ===========
- Here are the records from a busy telephone system with one million users:
- ┌─────────────────┬─────────┬─────────┐
- │RecNr │ Caller │ Called │
- ├─────────────────┼─────────┼─────────┤
- │ 1 │ 200007 │ 100053 │
- ├─────────────────┼─────────┼─────────┤
- │ 2 │ 600183 │ 500439 │
- ├─────────────────┼─────────┼─────────┤
- │ 3 │ 600863 │ 701497 │
- ├─────────────────┼─────────┼─────────┤
- │ ... │ ... │ ... │
- └─────────────────┴─────────┴─────────┘
- The telephone number of the caller and the called number in record n are
- Caller(n) = S[2n-1] and Called(n) = S[2n] where S[1,2,3,...] come from the
- "Lagged Fibonacci Generator":
- For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3] (modulo 1000000)
- For 56 ≤ k, S[k] = [S[k-24] + S[k-55]] (modulo 1000000)
- If Caller(n) = Called(n) then the user is assumed to have misdialled and
- the call fails; otherwise the call is successful.
- From the start of the records, we say that any pair of users X and Y are
- friends if X calls Y or vice-versa. Similarly, X is a friend of a friend
- of Z if X is a friend of Y and Y is a friend of Z; and so on for longer
- chains.
- The Prime Minister's phone number is 524287. After how many successful
- calls, not counting misdials, will 99% of the users (including the PM) be
- a friend, or a friend of a friend etc., of the Prime Minister?
- Answer: 2325629
- Problem 187
- ===========
- A composite is a number containing at least two prime factors. For
- example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3.
- There are ten composites below thirty containing precisely two, not
- necessarily distinct, prime factors:4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
- How many composite integers, n < 10^8, have precisely two, not necessarily
- distinct, prime factors?
- Answer: 17427258
- Problem 188
- ===========
- The hyperexponentiation or tetration of a number a by a positive integer
- b, denoted by a↑↑b or ^ba, is recursively defined by:
- a↑↑1 = a,
- a↑↑(k+1) = a^(a↑↑k).
- Thus we have e.g. 3↑↑2 = 3^3 = 27, hence 3↑↑3 = 3^27 = 7625597484987 and
- 3↑↑4 is roughly 10^3.6383346400240996*10^12.
- Find the last 8 digits of 1777↑↑1855.
- Answer: 95962097
- Problem 189
- ===========
- Consider the following configuration of 64 triangles:
- We wish to colour the interior of each triangle with one of three colours:
- red, green or blue, so that no two neighbouring triangles have the same
- colour. Such a colouring shall be called valid. Here, two triangles are
- said to be neighbouring if they share an edge.
- Note: if they only share a vertex, then they are not neighbours.
- For example, here is a valid colouring of the above grid:
- A colouring C' which is obtained from a colouring C by rotation or
- reflection is considered distinct from C unless the two are identical.
- How many distinct valid colourings are there for the above configuration?
- p_189_grid.gif
- p_189_colours.gif
- Answer: d3dfdd37601678212b746c34699f1484
- Problem 190
- ===========
- Let S[m] = (x[1], x[2], ... , x[m]) be the m-tuple of positive real
- numbers with x[1] + x[2] + ... + x[m] = m for which P[m] = x[1] * x[2]^2 *
- ... * x[m]^m is maximised.
- For example, it can be verified that [P[10]] = 4112 ([ ] is the integer
- part function).
- Find Σ[P[m]] for 2 ≤ m ≤ 15.
- Answer: 40cfcabd9b30d79ec81151fc756e9946
- Problem 191
- ===========
- A particular school offers cash rewards to children with good attendance
- and punctuality. If they are absent for three consecutive days or late on
- more than one occasion then they forfeit their prize.
- During an n-day period a trinary string is formed for each child
- consisting of L's (late), O's (on time), and A's (absent).
- Although there are eighty-one trinary strings for a 4-day period that can
- be formed, exactly forty-three strings would lead to a prize:
- OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA
- OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO
- AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL
- AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA
- LAOO LAOA LAAO
- How many "prize" strings exist over a 30-day period?
- Answer: e04dfa598b22a87570f63063f3ff595d
- Problem 192
- ===========
- Let x be a real number.
- A best approximation to x for the denominator bound d is a rational number
- r/s in reduced form, with s ≤ d, such that any rational number which is
- closer to x than r/s has a denominator larger than d:
- |p/q-x| < |r/s-x| ⇒ q > d
- For example, the best approximation to √13 for the denominator bound 20 is
- 18/5 and the best approximation to √13 for the denominator bound 30 is
- 101/28.
- Find the sum of all denominators of the best approximations to √n for the
- denominator bound 10^12, where n is not a perfect square and 1 < n ≤
- 100000.
- Answer: e5ec7d4b094709b1fcebbd73b10e6264
- Problem 193
- ===========
- A positive integer n is called squarefree, if no square of a prime divides
- n, thus 1, 2, 3, 5, 6, 7, 10, 11 are squarefree, but not 4, 8, 9, 12.
- How many squarefree numbers are there below 2^50?
- Answer: ea29fcf755b560777b0b6d8714234d18
- Problem 194
- ===========
- Consider graphs built with the units A: and B: , where the units are glued
- alongthe vertical edges as in the graph .
- A configuration of type (a,b,c) is a graph thus built of a units A and b
- units B, where the graph's vertices are coloured using up to c colours, so
- that no two adjacent vertices have the same colour.
- The compound graph above is an example of a configuration of type (2,2,6),
- in fact of type (2,2,c) for all c ≥ 4.
- Let N(a,b,c) be the number of configurations of type (a,b,c).
- For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736.
- Find the last 8 digits of N(25,75,1984).
- p_194_GraphA.png
- p_194_GraphB.png
- p_194_Fig.png
- Answer: 61190912
- Problem 195
- ===========
- Let's call an integer sided triangle with exactly one angle of 60 degrees
- a 60-degree triangle.
- Let r be the radius of the inscribed circle of such a 60-degree triangle.
- There are 1234 60-degree triangles for which r ≤ 100.
- Let T(n) be the number of 60-degree triangles for which r ≤ n, so
- T(100) = 1234, T(1000) = 22767, and T(10000) = 359912.
- Find T(1053779).
- Answer: 75085391
- Problem 196
- ===========
- Build a triangle from all positive integers in the following way:
- 1
- 2 3
- 4 5 6
- 7 8 9 10
- 11 12 13 14 15
- 16 17 18 19 20 21
- 22 23 24 25 26 27 28
- 29 30 31 32 33 34 35 36
- 37 38 39 40 41 42 43 44 45
- 46 47 48 49 50 51 52 53 54 55
- 56 57 58 59 60 61 62 63 64 65 66
- . . .
- Each positive integer has up to eight neighbours in the triangle.
- A set of three primes is called a prime triplet if one of the three primes
- has the other two as neighbours in the triangle.
- For example, in the second row, the prime numbers 2 and 3 are elements of
- some prime triplet.
- If row 8 is considered, it contains two primes which are elements of some
- prime triplet, i.e. 29 and 31.
- If row 9 is considered, it contains only one prime which is an element of
- some prime triplet: 37.
- Define S(n) as the sum of the primes in row n which are elements of any
- prime triplet.
- Then S(8)=60 and S(9)=37.
- You are given that S(10000)=950007619.
- Find S(5678027) + S(7208785).
- Answer: fb6b6b0a4b7b31ba429152bc0b6bd037
- Problem 197
- ===========
- Given is the function f(x) = ⌊2^30.403243784-x^2⌋ × 10^-9 ( ⌊ ⌋ is the
- floor-function),
- the sequence u[n] is defined by u[0] = -1 and u[n+1] = f(u[n]).
- Find u[n] + u[n+1] for n = 10^12.
- Give your answer with 9 digits after the decimal point.
- Answer: c98cbf87636906f2465d481be815e454
- Problem 198
- ===========
- A best approximation to a real number x for the denominator bound d is a
- rational number r/s (in reduced form) with s ≤ d, so that any rational
- number p/q which is closer to x than r/s has q > d.
- Usually the best approximation to a real number is uniquely determined for
- all denominator bounds. However, there are some exceptions, e.g. 9/40 has
- the two best approximations 1/4 and 1/5 for the denominator bound 6.We
- shall call a real number x ambiguous, if there is at least one denominator
- bound for which x possesses two best approximations. Clearly, an ambiguous
- number is necessarily rational.
- How many ambiguous numbers x = p/q,0 < x < 1/100, are there whose
- denominator q does not exceed 10^8?
- Answer: 52374425
- Problem 199
- ===========
- Three circles of equal radius are placed inside a larger circle such that
- each pair of circles is tangent to one another and the inner circles do
- not overlap. There are four uncovered "gaps" which are to be filled
- iteratively with more tangent circles.
- At each iteration, a maximally sized circle is placed in each gap, which
- creates more gaps for the next iteration. After 3 iterations (pictured),
- there are 108 gaps and the fraction of the area which is not covered by
- circles is 0.06790342, rounded to eight decimal places.
- What fraction of the area is not covered by circles after 10 iterations?
- Give your answer rounded to eight decimal places using the format
- x.xxxxxxxx .
- p_199_circles_in_circles.gif
- Answer: 0f8fd87159c28ae5fea6ac91a95d48dd
- Problem 200
- ===========
- We shall define a sqube to be a number of the form, p^2q^3, where p and q
- are distinct primes.
- For example, 200 = 5^22^3 or 120072949 = 23^261^3.
- The first five squbes are 72, 108, 200, 392, and 500.
- Interestingly, 200 is also the first number for which you cannot change
- any single digit to make a prime; we shall call such numbers, prime-proof.
- The next prime-proof sqube which contains the contiguous sub-string "200"
- is 1992008.
- Find the 200th prime-proof sqube containing the contiguous sub-string
- "200".
- Answer: c911c8e346aa813da5f5ed4f8e9128d8
- Problem 201
- ===========
- For any set A of numbers, let sum(A) be the sum of the elements of A.
- Consider the set B = {1,3,6,8,10,11}.
- There are 20 subsets of B containing three elements, and their sums are:
- sum({1,3,6}) = 10,
- sum({1,3,8}) = 12,
- sum({1,3,10}) = 14,
- sum({1,3,11}) = 15,
- sum({1,6,8}) = 15,
- sum({1,6,10}) = 17,
- sum({1,6,11}) = 18,
- sum({1,8,10}) = 19,
- sum({1,8,11}) = 20,
- sum({1,10,11}) = 22,
- sum({3,6,8}) = 17,
- sum({3,6,10}) = 19,
- sum({3,6,11}) = 20,
- sum({3,8,10}) = 21,
- sum({3,8,11}) = 22,
- sum({3,10,11}) = 24,
- sum({6,8,10}) = 24,
- sum({6,8,11}) = 25,
- sum({6,10,11}) = 27,
- sum({8,10,11}) = 29.
- Some of these sums occur more than once, others are unique.
- For a set A, let U(A,k) be the set of unique sums of k-element subsets of
- A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and
- sum(U(B,3)) = 156.
- Now consider the 100-element set S = {1^2, 2^2, ... , 100^2}.
- S has 100891344545564193334812497256 50-element subsets.
- Determine the sum of all integers which are the sum of exactly one of the
- 50-element subsets of S, i.e. find sum(U(S,50)).
- Answer: 115039000
- Problem 202
- ===========
- Three mirrors are arranged in the shape of an equilateral triangle, with
- their reflective surfaces pointing inwards. There is an infinitesimal gap
- at each vertex of the triangle through which a laser beam may pass.
- Label the vertices A, B and C. There are 2 ways in which a laser beam may
- enter vertex C, bounce off 11 surfaces, then exit through the same vertex:
- one way is shown below; the other is the reverse of that.
- There are 80840 ways in which a laser beam may enter vertex C, bounce off
- 1000001 surfaces, then exit through the same vertex.
- In how many ways can a laser beam enter at vertex C, bounce off
- 12017639147 surfaces, then exit through the same vertex?
- p_201_laserbeam.gif
- Answer: e9774949b5efad0d40d60ede379c5321
- Problem 203
- ===========
- The binomial coefficients ^nC[k] can be arranged in triangular form,
- Pascal's triangle, like this:
- 1
- 1 1
- 1 2 1
- 1 3 3 1
- 1 4 6 4 1
- 1 5 10 10 5 1
- 1 6 15 20 15 6 1
- 1 7 21 35 35 21 7 1
- .........
- It can be seen that the first eight rows of Pascal's triangle contain
- twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35.
- A positive integer n is called squarefree if no square of a prime divides
- n.Of the twelve distinct numbers in the first eight rows of Pascal's
- triangle, all except 4 and 20 are squarefree.The sum of the distinct
- squarefree numbers in the first eight rows is 105.
- Find the sum of the distinct squarefree numbers in the first 51 rows of
- Pascal's triangle.
- Answer: d7ec16d216c923d3c927f46cfc914e92
- Problem 204
- ===========
- A Hamming number is a positive number which has no prime factor larger
- than 5.
- So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.
- There are 1105 Hamming numbers not exceeding 10^8.
- We will call a positive number a generalised Hamming number of type n, if
- it has no prime factor larger than n.
- Hence the Hamming numbers are the generalised Hamming numbers of type 5.
- How many generalised Hamming numbers of type 100 are there which don't
- exceed 10^9?
- Answer: 2944730
- Problem 205
- ===========
- Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2,
- 3, 4.
- Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4,
- 5, 6.
- Peter and Colin roll their dice and compare totals: the highest total
- wins. The result is a draw if the totals are equal.
- What is the probability that Pyramidal Pete beats Cubic Colin? Give your
- answer rounded to seven decimal places in the form 0.abcdefg
- Answer: ba6c6c3888227a0799eca38191b587be
- Problem 206
- ===========
- Find the unique positive integer whose square has the form
- 1_2_3_4_5_6_7_8_9_0,
- where each “_” is a single digit.
- Answer: 09f9d87cb4b1ebb34e1f607e55a351d8
- Problem 207
- ===========
- For some positive integers k, there exists an integer partition of the
- form 4^t = 2^t + k,
- where 4^t, 2^t, and k are all positive integers and t is a real number.
- The first two such partitions are 4^1 = 2^1 + 2 and 4^1.5849625... =
- 2^1.5849625... + 6.
- Partitions where t is also an integer are called perfect.
- For any m ≥ 1 let P(m) be the proportion of such partitions that are
- perfect with k ≤ m.
- Thus P(6) = 1/2.
- In the following table are listed some values of P(m)
- P(5) = 1/1
- P(10) = 1/2
- P(15) = 2/3
- P(20) = 1/2
- P(25) = 1/2
- P(30) = 2/5
- ...
- P(180) = 1/4
- P(185) = 3/13
- Find the smallest m for which P(m) < 1/12345
- Answer: 3f17b264ed1717fe5fbde1e399bd501f
- Problem 208
- ===========
- A robot moves in a series of one-fifth circular arcs (72°), with a free
- choice of a clockwise or an anticlockwise arc for each step, but no
- turning on the spot.
- One of 70932 possible closed paths of 25 arcs starting northward is
- Given that the robot starts facing North, how many journeys of 70 arcs in
- length can it take that return it, after the final arc, to its starting
- position?
- (Any arc may be traversed multiple times.)
- p_208_robotwalk.gif
- Answer: 3010e33173f30e0aac79e84835b48823
- Problem 209
- ===========
- A k-input binary truth table is a map from k input bits(binary digits, 0
- [false] or 1 [true]) to 1 output bit. For example, the 2-input binary
- truth tables for the logical AND and XOR functions are:
- ┌────┬────┬─────────┐
- │ x │ y │x AND y │
- ├────┼────┼─────────┤
- │ 0 │ 0 │ 0 │
- ├────┼────┼─────────┤
- │ 0 │ 1 │ 0 │
- ├────┼────┼─────────┤
- │ 1 │ 0 │ 0 │
- ├────┼────┼─────────┤
- │ 1 │ 1 │ 1 │
- └────┴────┴─────────┘
- ┌────┬────┬─────────┐
- │ x │ y │x XOR y │
- ├────┼────┼─────────┤
- │ 0 │ 0 │ 0 │
- ├────┼────┼─────────┤
- │ 0 │ 1 │ 1 │
- ├────┼────┼─────────┤
- │ 1 │ 0 │ 1 │
- ├────┼────┼─────────┤
- │ 1 │ 1 │ 0 │
- └────┴────┴─────────┘
- How many 6-input binary truth tables, τ, satisfy the formula
- τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0
- Answer: 954157aa4762df2ee29580ee5a351b13
- Problem 210
- ===========
- Consider the set S(r) of points (x,y) with integer coordinates satisfying
- |x| + |y| ≤ r.
- Let O be the point (0,0) and C the point (r/4,r/4).
- Let N(r) be the number of points B in S(r), so that the triangle OBC has
- an obtuse angle, i.e. the largest angle α satisfies 90°<α<180°.
- So, for example, N(4)=24 and N(8)=100.
- What is N(1,000,000,000)?
- Answer: 0c808b02789c4db462322ab2ac070bbb
- Problem 211
- ===========
- For a positive integer n, let σ[2](n) be the sum of the squares of its
- divisors. For example,
- σ[2](10) = 1 + 4 + 25 + 100 = 130.
- Find the sum of all n, 0 < n < 64,000,000 such that σ[2](n) is a perfect
- square.
- Answer: 5fe0ed146690e7bca448687a94353a73
- Problem 212
- ===========
- An axis-aligned cuboid, specified by parameters { (x[0],y[0],z[0]),
- (dx,dy,dz) }, consists of all points (X,Y,Z) such that x[0] ≤ X ≤ x[0]+dx,
- y[0] ≤ Y ≤ y[0]+dy and z[0] ≤ Z ≤ z[0]+dz. The volume of the cuboid is the
- product, dx × dy × dz. The combined volume of a collection of cuboids is
- the volume of their union and will be less than the sum of the individual
- volumes if any cuboids overlap.
- Let C[1],...,C[50000] be a collection of 50000 axis-aligned cuboids such
- that C[n] has parameters
- x[0] = S[6n-5] modulo 10000
- y[0] = S[6n-4] modulo 10000
- z[0] = S[6n-3] modulo 10000
- dx = 1 + (S[6n-2] modulo 399)
- dy = 1 + (S[6n-1] modulo 399)
- dz = 1 + (S[6n] modulo 399)
- where S[1],...,S[300000] come from the "Lagged Fibonacci Generator":
- For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3] (modulo 1000000)
- For 56 ≤ k, S[k] = [S[k-24] + S[k-55]] (modulo 1000000)
- Thus, C[1] has parameters {(7,53,183),(94,369,56)}, C[2] has parameters
- {(2383,3563,5079),(42,212,344)}, and so on.
- The combined volume of the first 100 cuboids, C[1],...,C[100], is
- 723581599.
- What is the combined volume of all 50000 cuboids, C[1],...,C[50000] ?
- Answer: 76650c9c077929e1ce5a80a1ac81fa96
- Problem 213
- ===========
- A 30×30 grid of squares contains 900 fleas, initially one flea per square.
- When a bell is rung, each flea jumps to an adjacent square at random
- (usually 4 possibilities, except for fleas on the edge of the grid or at
- the corners).
- What is the expected number of unoccupied squares after 50 rings of the
- bell? Give your answer rounded to six decimal places.
- Answer: f81ee7dd444a3d895a4a446f9d115bf8
- Problem 214
- ===========
- Let φ be Euler's totient function, i.e. for a natural number n,φ(n) is the
- number of k, 1 ≤ k ≤ n, for which gcd(k,n) = 1.
- By iterating φ, each positive integer generates a decreasing chain of
- numbers ending in 1.
- E.g. if we start with 5 the sequence 5,4,2,1 is generated.
- Here is a listing of all chains with length 4:
- 5,4,2,1
- 7,6,2,1
- 8,4,2,1
- 9,6,2,1
- 10,4,2,1
- 12,4,2,1
- 14,6,2,1
- 18,6,2,1
- Only two of these chains start with a prime, their sum is 12.
- What is the sum of all primes less than 40000000 which generate a chain of
- length 25?
- Answer: 1cefd865483c03552d5247ffb05685c7
- Problem 215
- ===========
- Consider the problem of building a wall out of 2×1 and 3×1 bricks
- (horizontal×vertical dimensions) such that, for extra strength, the gaps
- between horizontally-adjacent bricks never line up in consecutive layers,
- i.e. never form a "running crack".
- For example, the following 9×3 wall is not acceptable due to the running
- crack shown in red:
- There are eight ways of forming a crack-free 9×3 wall, written W(9,3) = 8.
- Calculate W(32,10).
- p_215_crackfree.gif
- Answer: 60212c9ec4a6cd1d14277c32b6adf2d8
- Problem 216
- ===========
- Consider numbers t(n) of the form t(n) = 2n^2-1 with n > 1.
- The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161.
- It turns out that only 49 = 7*7 and 161 = 7*23 are not prime.
- For n ≤ 10000 there are 2202 numbers t(n) that are prime.
- How many numbers t(n) are prime for n ≤ 50,000,000 ?
- Answer: 5437849
- Problem 217
- ===========
- A positive integer with k (decimal) digits is called balanced if its first
- ⌈^k/[2]⌉ digits sum to the same value as its last ⌈^k/[2]⌉ digits, where
- ⌈x⌉, pronounced ceiling of x, is the smallest integer ≥ x, thus ⌈π⌉ = 4
- and ⌈5⌉ = 5.
- So, for example, all palindromes are balanced, as is 13722.
- Let T(n) be the sum of all balanced numbers less than 10^n.
- Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890.
- Find T(47) mod 3^15
- Answer: 6273134
- Problem 218
- ===========
- Consider the right angled triangle with sides a=7, b=24 and c=25.The area
- of this triangle is 84, which is divisible by the perfect numbers 6 and
- 28.
- Moreover it is a primitive right angled triangle as gcd(a,b)=1 and
- gcd(b,c)=1.
- Also c is a perfect square.
- We will call a right angled triangle perfect if
- -it is a primitive right angled triangle
- -its hypotenuse is a perfect square
- We will call a right angled triangle super-perfect if
- -it is a perfect right angled triangle and
- -its area is a multiple of the perfect numbers 6 and 28.
- How many perfect right-angled triangles with c≤10^16 exist that are not
- super-perfect?
- Answer: 0
- Problem 219
- ===========
- Let A and B be bit strings (sequences of 0's and 1's).
- If A is equal to the leftmost length(A) bits of B, then A is said to be a
- prefix of B.
- For example, 00110 is a prefix of 001101001, but not of 00111 or 100110.
- A prefix-free code of size n is a collection of n distinct bit strings
- such that no string is a prefix of any other. For example, this is a
- prefix-free code of size 6:
- 0000, 0001, 001, 01, 10, 11
- Now suppose that it costs one penny to transmit a '0' bit, but four pence
- to transmit a '1'.
- Then the total cost of the prefix-free code shown above is 35 pence, which
- happens to be the cheapest possible for the skewed pricing scheme in
- question.
- In short, we write Cost(6) = 35.
- What is Cost(10^9) ?
- Answer: 578c22ef288b88c60fbcf4541351aff5
- Problem 220
- ===========
- Let D[0] be the two-letter string "Fa". For n≥1, derive D[n] from D[n-1]
- by the string-rewriting rules:
- "a" → "aRbFR"
- "b" → "LFaLb"
- Thus, D[0] = "Fa", D[1] = "FaRbFR", D[2] = "FaRbFRRLFaLbFR", and so on.
- These strings can be interpreted as instructions to a computer graphics
- program, with "F" meaning "draw forward one unit", "L" meaning "turn left
- 90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being
- ignored. The initial position of the computer cursor is (0,0), pointing up
- towards (0,1).
- Then D[n] is an exotic drawing known as the Heighway Dragon of order n.
- For example, D[10] is shown below; counting each "F" as one step, the
- highlighted spot at (18,16) is the position reached after 500 steps.
- What is the position of the cursor after 10^12 steps in D[50] ?
- Give your answer in the form x,y with no spaces.
- p_220.gif
- Answer: e2018d8efde8ea00319f1adc042f150b
- Problem 221
- ===========
- We shall call a positive integer A an "Alexandrian integer", if there
- exist integers p, q, r such that:
- A = p · q · r and 1 = 1 + 1 + 1
- A p q r
- For example, 630 is an Alexandrian integer (p = 5, q = −7, r = −18).In
- fact, 630 is the 6^th Alexandrian integer, the first 6 Alexandrian
- integers being: 6, 42, 120, 156, 420 and 630.
- Find the 150000^th Alexandrian integer.
- Answer: cb000c24f653d9c8f78b74123e6515ab
- Problem 222
- ===========
- What is the length of the shortest pipe, of internal radius 50mm, that can
- fully contain 21 balls of radii 30mm, 31mm, ..., 50mm?
- Give your answer in micrometres (10^-6 m) rounded to the nearest integer.
- Answer: 1590933
- Problem 223
- ===========
- Let us call an integer sided triangle with sides a ≤ b ≤ c barely acute if
- the sides satisfy
- a^2 + b^2 = c^2 + 1.
- How many barely acute triangles are there with perimeter ≤ 25,000,000?
- Answer: 61614848
- Problem 224
- ===========
- Let us call an integer sided triangle with sides a ≤ b ≤ c barely obtuse
- if the sides satisfy
- a^2 + b^2 = c^2 - 1.
- How many barely obtuse triangles are there with perimeter ≤ 75,000,000?
- Answer: 4137330
- Problem 225
- ===========
- The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ...
- is defined by T[1] = T[2] = T[3] = 1 and T[n] = T[n-1] + T[n-2] + T[n-3].
- It can be shown that 27 does not divide any terms of this sequence.
- In fact, 27 is the first odd number with this property.
- Find the 124^th odd number that does not divide any terms of the above
- sequence.
- Answer: 2009
- Problem 226
- ===========
- The blancmange curve is the set of points (x,y) such that 0 ≤ x ≤ 1 and ,
- where s(x) = the distance from x to the nearest integer.
- The area under the blancmange curve is equal to ½, shown in pink in the
- diagram below.
- [1]blancmange curve
- Let C be the circle with centre (¼,½) and radius ¼, shown in black in the
- diagram.
- What area under the blancmange curve is enclosed by C?
- Give your answer rounded to eight decimal places in the form 0.abcdefgh
- Visible links
- p_226_formula.gif
- p_226_scoop2.gif
- Answer: ce6fd32d1d2fb58c4c0c1f7962c39f04
- Problem 227
- ===========
- "The Chase" is a game played with two dice and an even number of players.
- The players sit around a table; the game begins with two opposite players
- having one die each. On each turn, the two players with a die roll it.
- If a player rolls a 1, he passes the die to his neighbour on the left; if
- he rolls a 6, he passes the die to his neighbour on the right; otherwise,
- he keeps the die for the next turn.
- The game ends when one player has both dice after they have been rolled
- and passed; that player has then lost.
- In a game with 100 players, what is the expected number of turns the game
- lasts?
- Give your answer rounded to ten significant digits.
- Answer: 7b87cd0a96f0f2f12f911cdc66608d95
- Problem 228
- ===========
- Let S[n] be the regular n-sided polygon – or shape – whose vertices v[k]
- (k = 1,2,…,n) have coordinates:
- x[k] = cos( ^2k-1/[n] ×180° )
- y[k] = sin( ^2k-1/[n] ×180° )
- Each S[n] is to be interpreted as a filled shape consisting of all points
- on the perimeter and in the interior.
- The Minkowski sum, S+T, of two shapes S and T is the result of adding
- every point in S to every point in T, where point addition is performed
- coordinate-wise: (u, v) + (x, y) = (u+x, v+y).
- For example, the sum of S[3] and S[4] is the six-sided shape shown in pink
- below:
- [1]picture showing S_3 + S_4
- How many sides does S[1864] + S[1865] + … + S[1909] have?
- Visible links
- p_228.png
- Answer: 86226
- Problem 229
- ===========
- Consider the number 3600. It is very special, because
- 3600 = 48^2 + 36^2
- 3600 = 20^2 + 2×40^2
- 3600 = 30^2 + 3×30^2
- 3600 = 45^2 + 7×15^2
- Similarly, we find that 88201 = 99^2 + 280^2 = 287^2 + 2×54^2 = 283^2 +
- 3×52^2 = 197^2 + 7×84^2.
- In 1747, Euler proved which numbers are representable as a sum of two
- squares.We are interested in the numbers n which admit representations of
- all of the following four types:
- n = a[1]^2 + b[1]^2
- n = a[2]^2 + 2 b[2]^2
- n = a[3]^2 + 3 b[3]^2
- n = a[7]^2 + 7 b[7]^2,
- where the a[k] and b[k] are positive integers.
- There are 75373 such numbers that do not exceed 10^7.
- How many such numbers are there that do not exceed 2×10^9?
- Answer: 11325263
- Problem 230
- ===========
- For any two strings of digits, A and B, we define F[A,B] to be the
- sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of
- the previous two.
- Further, we define D[A,B](n) to be the n^th digit in the first term of
- F[A,B] that contains at least n digits.
- Example:
- Let A=1415926535, B=8979323846. We wish to find D[A,B](35), say.
- The first few terms of F[A,B] are:
- 1415926535
- 8979323846
- 14159265358979323846
- 897932384614159265358979323846
- 14159265358979323846897932384614159265358979323846
- Then D[A,B](35) is the 35^th digit in the fifth term, which is 9.
- Now we use for A the first 100 digits of π behind the decimal point:
- 14159265358979323846264338327950288419716939937510
- 58209749445923078164062862089986280348253421170679
- and for B the next hundred digits:
- 82148086513282306647093844609550582231725359408128
- 48111745028410270193852110555964462294895493038196 .
- Find ∑[n = 0,1,...,17] 10^n× D[A,B]((127+19n)×7^n) .
- Answer: 040735038021ff4704bbd3a0964369ef
- Problem 231
- ===========
- The binomial coefficient ^10C[3] = 120.
- 120 = 2^3 × 3 × 5 = 2 × 2 × 2 × 3 × 5, and 2 + 2 + 2 + 3 + 5 = 14.
- So the sum of the terms in the prime factorisation of ^10C[3] is 14.
- Find the sum of the terms in the prime factorisation of
- ^20000000C[15000000].
- Answer: ef8bc4d9a843e71126bd10b5065132a5
- Problem 232
- ===========
- Two players share an unbiased coin and take it in turns to play "The
- Race". On Player 1's turn, he tosses the coin once: if it comes up Heads,
- he scores one point; if it comes up Tails, he scores nothing. On Player
- 2's turn, she chooses a positive integer T and tosses the coin T times: if
- it comes up all Heads, she scores 2^T-1 points; otherwise, she scores
- nothing. Player 1 goes first. The winner is the first to 100 or more
- points.
- On each turn Player 2 selects the number, T, of coin tosses that maximises
- the probability of her winning.
- What is the probability that Player 2 wins?
- Give your answer rounded to eight decimal places in the form 0.abcdefgh .
- Answer: c8d5b243aa6e6b507725766f7c197a1d
- Problem 233
- ===========
- Let f(N) be the number of points with integer coordinates that are on a
- circle passing through (0,0), (N,0),(0,N), and (N,N).
- It can be shown that f(10000) = 36.
- What is the sum of all positive integers N ≤ 10^11 such that f(N) = 420 ?
- Answer: 7e80b27798170abb493e3b4671bd82ca
- Problem 234
- ===========
- For an integer n ≥ 4, we define the lower prime square root of n, denoted
- by lps(n), as the largest prime ≤ √n and the upper prime square root of n,
- ups(n), as the smallest prime ≥ √n.
- So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37.
- Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n)
- divides n, but not both.
- The sum of the semidivisible numbers not exceeding 15 is 30, the numbers
- are 8, 10 and 12.
- 15 is not semidivisible because it is a multiple of both lps(15) = 3 and
- ups(15) = 5.
- As a further example, the sum of the 92 semidivisible numbers up to 1000
- is 34825.
- What is the sum of all semidivisible numbers not exceeding 999966663333 ?
- Answer: c24a5d60f8ce5d04dec7466987c84d68
- Problem 235
- ===========
- Given is the arithmetic-geometric sequence u(k) = (900-3k)r^k-1.
- Let s(n) = Σ[k=1...n]u(k).
- Find the value of r for which s(5000) = -600,000,000,000.
- Give your answer rounded to 12 places behind the decimal point.
- Answer: 41b13508789be1001308e065d4f83ea2
- Problem 236
- ===========
- Suppliers 'A' and 'B' provided the following numbers of products for the
- luxury hamper market:
- Product 'A' 'B'
- Beluga Caviar 5248 640
- Christmas Cake 1312 1888
- Gammon Joint 2624 3776
- Vintage Port 5760 3776
- Champagne Truffles 3936 5664
- Although the suppliers try very hard to ship their goods in perfect
- condition, there is inevitably some spoilage - i.e. products gone bad.
- The suppliers compare their performance using two types of statistic:
- • The five per-product spoilage rates for each supplier are equal to the
- number of products gone bad divided by the number of products
- supplied, for each of the five products in turn.
- • The overall spoilage rate for each supplier is equal to the total
- number of products gone bad divided by the total number of products
- provided by that supplier.
- To their surprise, the suppliers found that each of the five per-product
- spoilage rates was worse (higher) for 'B' than for 'A' by the same factor
- (ratio of spoilage rates), m>1; and yet, paradoxically, the overall
- spoilage rate was worse for 'A' than for 'B', also by a factor of m.
- There are thirty-five m>1 for which this surprising result could have
- occurred, the smallest of which is 1476/1475.
- What's the largest possible value of m?
- Give your answer as a fraction reduced to its lowest terms, in the form
- u/v.
- Answer: 6e707fcffc510520d981ae16a29579bb
- Problem 237
- ===========
- Let T(n) be the number of tours over a 4 × n playing board such that:
- • The tour starts in the top left corner.
- • The tour consists of moves that are up, down, left, or right one
- square.
- • The tour visits each square exactly once.
- • The tour ends in the bottom left corner.
- The diagram shows one tour over a 4 × 10 board:
- T(10) is 2329. What is T(10^12) modulo 10^8?
- p_237.gif
- Answer: 15836928
- Problem 238
- ===========
- Create a sequence of numbers using the "Blum Blum Shub" pseudo-random
- number generator:
- s[0] = 14025256
- s[n+1] = s[n]^2 mod 20300713
- Concatenate these numbers s[0]s[1]s[2]… to create a string w of infinite
- length.
- Then, w = 14025256741014958470038053646…
- For a positive integer k, if no substring of w exists with a sum of digits
- equal to k, p(k) is defined to be zero. If at least one substring of w
- exists with a sum of digits equal to k, we define p(k) = z, where z is the
- starting position of the earliest such substring.
- For instance:
- The substrings 1, 14, 1402, …
- with respective sums of digits equal to 1, 5, 7, …
- start at position 1, hence p(1) = p(5) = p(7) = … = 1.
- The substrings 4, 402, 4025, …
- with respective sums of digits equal to 4, 6, 11, …
- start at position 2, hence p(4) = p(6) = p(11) = … = 2.
- The substrings 02, 0252, …
- with respective sums of digits equal to 2, 9, …
- start at position 3, hence p(2) = p(9) = … = 3.
- Note that substring 025 starting at position 3, has a sum of digits equal
- to 7, but there was an earlier substring (starting at position 1) with a
- sum of digits equal to 7, so p(7) = 1, not 3.
- We can verify that, for 0 < k ≤ 10^3, ∑ p(k) = 4742.
- Find ∑ p(k), for 0 < k ≤ 2·10^15.
- Answer: 424ed6613a372ccb9a90dddb8961ca16
- Problem 239
- ===========
- A set of disks numbered 1 through 100 are placed in a line in random
- order.
- What is the probability that we have a partial derangement such that
- exactly 22 prime number discs are found away from their natural positions?
- (Any number of non-prime disks may also be found in or out of their
- natural positions.)
- Give your answer rounded to 12 places behind the decimal point in the form
- 0.abcdefghijkl.
- Answer: 451fd2b8c19fbfec650a5c4699f6ef6e
- Problem 240
- ===========
- There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can
- be rolled so that the top three sum to 15. Some examples are:
- D[1],D[2],D[3],D[4],D[5] = 4,3,6,3,5
- D[1],D[2],D[3],D[4],D[5] = 4,3,3,5,6
- D[1],D[2],D[3],D[4],D[5] = 3,3,3,6,6
- D[1],D[2],D[3],D[4],D[5] = 6,6,3,3,3
- In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be
- rolled so that the top ten sum to 70?
- Answer: cb31a3106db3876e77cd160664cd683e
- Problem 241
- ===========
- For a positive integer n, let σ(n) be the sum of all divisors of n, so
- e.g. σ(6) = 1 + 2 + 3 + 6 = 12.
- A perfect number, as you probably know, is a number with σ(n) = 2n.
- Let us define the perfection quotient of a positive integer p(n) = σ(n) .
- as n
- Find the sum of all positive integers n ≤ 10^18 for which p(n) has the
- form k + ^1⁄[2], where k is an integer.
- Answer: 556bfef2cacd1eff8af9126c5c13dcbc
- Problem 242
- ===========
- Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element
- subsets with an odd sum of elements. For example, f(5,3) = 4, since the
- set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements,
- i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}.
- When all three values n, k and f(n,k) are odd, we say that they make
- an odd-triplet [n,k,f(n,k)].
- There are exactly five odd-triplets with n ≤ 10, namely:
- [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
- [9,9,f(9,9) = 1].
- How many odd-triplets are there with n ≤ 10^12 ?
- Answer: ba73cb75365ddca8f94a23e3fedfb6de
- Problem 243
- ===========
- A positive fraction whose numerator is less than its denominator is called
- a proper fraction.
- For any denominator, d, there will be d−1 proper fractions; for example,
- with d = 12:
- 1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 ,
- 8/12 , 9/12 , 10/12 , 11/12 .
- We shall call a fraction that cannot be cancelled down a resilient
- fraction.
- Furthermore we shall define the resilience of a denominator, R(d), to be
- the ratio of its proper fractions that are resilient; for example, R(12) =
- 4/11 .
- In fact, d = 12 is the smallest denominator having a resilience R(d) <
- 4/10 .
- Find the smallest denominator d, having a resilience R(d) < 15499/94744
- .
- Answer: 531721a10786c5c2a444b474fcf039f9
- Problem 244
- ===========
- You probably know the game Fifteen Puzzle. Here, instead of numbered
- tiles, we have seven red tiles and eight blue tiles.
- A move is denoted by the uppercase initial of the direction (Left, Right,
- Up, Down) in which the tile is slid, e.g. starting from configuration (S),
- by the sequence LULUR we reach the configuration (E):
- (S) , (E)
- For each path, its checksum is calculated by (pseudocode):
- checksum = 0
- checksum = (checksum × 243 + m[1]) mod 100 000 007
- checksum = (checksum × 243 + m[2]) mod 100 000 007
- …
- checksum = (checksum × 243 + m[n]) mod 100 000 007
- where m[k] is the ASCII value of the k^th letter in the move sequence and
- the ASCII values for the moves are:
- ┌──────┬─────┐
- │L │76 │
- ├──────┼─────┤
- │R │82 │
- ├──────┼─────┤
- │U │85 │
- ├──────┼─────┤
- │D │68 │
- └──────┴─────┘
- For the sequence LULUR given above, the checksum would be 19761398.
- Now, starting from configuration (S),find all shortest ways to reach
- configuration (T).
- (S) , (T)
- What is the sum of all checksums for the paths having the minimal length?
- p_244_start.gif
- p_244_example.gif
- p_244_start.gif
- p_244_target.gif
- Answer: 96356848
- Problem 245
- ===========
- We shall call a fraction that cannot be cancelled down a resilient
- fraction.
- Furthermore we shall define the resilience of a denominator, R(d), to be
- the ratio of its proper fractions that are resilient; for example, R(12) =
- ^4⁄[11].
- The resilience of a number d > 1 is φ(d) , where φ is Euler's totient
- then d - 1 function.
- We further define the coresilience of a number n > 1 as C(n) = n - φ(n) .
- n - 1
- The coresilience of a prime p is C(p) = 1 .
- p - 1
- Find the sum of all composite integers 1 < n ≤ 2×10^11, for which C(n) is
- a unit fraction.
- Answer: 0ebeb502fb0bd7157609835d27c266bc
- Problem 246
- ===========
- A definition for an ellipse is:
- Given a circle c with centre M and radius r and a point G such that
- d(G,M)<r, the locus of the points that are equidistant from c and G form
- an ellipse.
- The construction of the points of the ellipse is shown below.
- Given are the points M(-2000,1500) and G(8000,1500).
- Given is also the circle c with centre M and radius 15000.
- The locus of the points that are equidistant from G and c form an ellipse
- e.
- From a point P outside e the two tangents t[1] and t[2] to the ellipse are
- drawn.
- Let the points where t[1] and t[2] touch the ellipse be R and S.
- For how many lattice points P is angle RPS greater than 45 degrees?
- p_246_anim.gif
- p_246_ellipse.gif
- Answer: 94c521ffeb906391d161b66fec433827
- Problem 247
- ===========
- Consider the region constrained by 1 ≤ x and 0 ≤ y ≤ ^1/[x].
- Let S[1] be the largest square that can fit under the curve.
- Let S[2] be the largest square that fits in the remaining area, and so on.
- Let the index of S[n] be the pair (left, below) indicating the number of
- squares to the left of S[n] and the number of squares below S[n].
- The diagram shows some such squares labelled by number.
- S[2] has one square to its left and none below, so the index of S[2] is
- (1,0).
- It can be seen that the index of S[32] is (1,1) as is the index of S[50].
- 50 is the largest n for which the index of S[n] is (1,1).
- What is the largest n for which the index of S[n] is (3,3)?
- p_247_hypersquares.gif
- Answer: 782252
- Problem 248
- ===========
- The first number n for which φ(n)=13! is 6227180929.
- Find the 150,000^th such number.
- Answer: b69a3ba674f6c7c5f2ce244f9e9cc873
- Problem 249
- ===========
- Let S = {2, 3, 5, ..., 4999} be the set of prime numbers less than 5000.
- Find the number of subsets of S, the sum of whose elements is a prime
- number.
- Enter the rightmost 16 digits as your answer.
- Answer: a470ee3ca52f2b68d7034e48b39b8b26
- Problem 250
- ===========
- Find the number of non-empty subsets of {1^1, 2^2, 3^3,...,
- 250250^250250}, the sum of whose elements is divisible by 250. Enter the
- rightmost 16 digits as your answer.
- Answer: 4a5614f3700956273fe0d271f921d5f4
- Problem 251
- ===========
- A triplet of positive integers (a,b,c) is called a Cardano Triplet if it
- satisfies the condition:
- For example, (2,1,5) is a Cardano Triplet.
- There exist 149 Cardano Triplets for which a+b+c ≤ 1000.
- Find how many Cardano Triplets exist such that a+b+c ≤ 110,000,000.
- p_251_cardano.gif
- Answer: 18946051
- Problem 252
- ===========
- Given a set of points on a plane, we define a convex hole to be a convex
- polygon having as vertices any of the given points and not containing any
- of the given points in its interior (in addition to the vertices, other
- given points may lie on the perimeter of the polygon).
- As an example, the image below shows a set of twenty points and a few such
- convex holes. The convex hole shown as a red heptagon has an area equal to
- 1049694.5 square units, which is the highest possible area for a convex
- hole on the given set of points.
- For our example, we used the first 20 points (T[2k−1], T[2k]), for
- k = 1,2,…,20, produced with the pseudo-random number generator:
- S[0] =[ ] 290797[ ]
- S[n+1] =[ ] S[n]^2 mod 50515093
- T[n] =[ ] ( S[n] mod 2000 ) − 1000^
- i.e. (527, 144), (−488, 732), (−454, −947), …
- What is the maximum area for a convex hole on the set containing the first
- 500 points in the pseudo-random sequence?
- Specify your answer including one digit after the decimal point.
- p_252_convexhole.gif
- Answer: 53b1ced82e1b588d756750c4d2f77e0d
- Problem 253
- ===========
- A small child has a “number caterpillar” consisting of forty jigsaw
- pieces, each with one number on it, which, when connected together in a
- line, reveal the numbers 1 to 40 in order.
- Every night, the child's father has to pick up the pieces of the
- caterpillar that have been scattered across the play room. He picks up the
- pieces at random and places them in the correct order.
- As the caterpillar is built up in this way, it forms distinct segments
- that gradually merge together.
- The number of segments starts at zero (no pieces placed), generally
- increases up to about eleven or twelve, then tends to drop again before
- finishing at a single segment (all pieces placed).
- For example:
- ┌────────────┬───────────────┐
- │Piece Placed│Segments So Far│
- ├────────────┼───────────────┤
- │ 12 │ 1 │
- ├────────────┼───────────────┤
- │ 4 │ 2 │
- ├────────────┼───────────────┤
- │ 29 │ 3 │
- ├────────────┼───────────────┤
- │ 6 │ 4 │
- ├────────────┼───────────────┤
- │ 34 │ 5 │
- ├────────────┼───────────────┤
- │ 5 │ 4 │
- ├────────────┼───────────────┤
- │ 35 │ 4 │
- ├────────────┼───────────────┤
- │ … │ … │
- └────────────┴───────────────┘
- Let M be the maximum number of segments encountered during a random
- tidy-up of the caterpillar.
- For a caterpillar of ten pieces, the number of possibilities for each M is
- ┌────────┬─────────────┐
- │ M │Possibilities│
- ├────────┼─────────────┤
- │ 1 │ 512 │
- ├────────┼─────────────┤
- │ 2 │ 250912 │
- ├────────┼─────────────┤
- │ 3 │1815264 │
- ├────────┼─────────────┤
- │ 4 │1418112 │
- ├────────┼─────────────┤
- │ 5 │ 144000 │
- └────────┴─────────────┘
- so the most likely value of M is 3 and the average value is
- ^385643⁄[113400] = 3.400732, rounded to six decimal places.
- The most likely value of M for a forty-piece caterpillar is 11; but what
- is the average value of M?
- Give your answer rounded to six decimal places.
- Answer: 228de0a37019fd7c7051029f3d126422
- Problem 254
- ===========
- Define f(n) as the sum of the factorials of the digits of n. For example,
- f(342) = 3! + 4! + 2! = 32.
- Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5.
- Define g(i) to be the smallest positive integer n such that sf(n) = i.
- Though sf(342) is 5, sf(25) is also 5, and it can be verified that g(5) is
- 25.
- Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7.
- Further, it can be verified that g(20) is 267 and ∑ sg(i) for 1 ≤ i ≤ 20
- is 156.
- What is ∑ sg(i) for 1 ≤ i ≤ 150?
- Answer: 936014adf2de65d41979ad900325e485
- Problem 255
- ===========
- We define the rounded-square-root of a positive integer n as the square
- root of n rounded to the nearest integer.
- The following procedure (essentially Heron's method adapted to integer
- arithmetic) finds the rounded-square-root of n:
- Let d be the number of digits of the number n.
- If d is odd, set x[0] = 2×10^(d-1)⁄2.
- If d is even, set x[0] = 7×10^(d-2)⁄2.
- Repeat:
- until x[k+1] = x[k].
- As an example, let us find the rounded-square-root of n = 4321.
- n has 4 digits, so x[0] = 7×10^(4-2)⁄2 = 70.
- Since x[2] = x[1], we stop here.
- So, after just two iterations, we have found that the rounded-square-root
- of 4321 is 66 (the actual square root is 65.7343137…).
- The number of iterations required when using this method is surprisingly
- low.
- For example, we can find the rounded-square-root of a 5-digit integer
- (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the
- average value was rounded to 10 decimal places).
- Using the procedure described above, what is the average number of
- iterations required to find the rounded-square-root of a 14-digit number
- (10^13 ≤ n < 10^14)?
- Give your answer rounded to 10 decimal places.
- Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling
- function respectively.
- p_255_Heron.gif
- p_255_Example.gif
- Answer: 12be028b156b49faa1137febda940ab5
- Problem 256
- ===========
- Tatami are rectangular mats, used to completely cover the floor of a room,
- without overlap.
- Assuming that the only type of available tatami has dimensions 1×2, there
- are obviously some limitations for the shape and size of the rooms that
- can be covered.
- For this problem, we consider only rectangular rooms with integer
- dimensions a, b and even size s = a·b.
- We use the term 'size' to denote the floor surface area of the room, and —
- without loss of generality — we add the condition a ≤ b.
- There is one rule to follow when laying out tatami: there must be no
- points where corners of four different mats meet.
- For example, consider the two arrangements below for a 4×4 room:
- The arrangement on the left is acceptable, whereas the one on the right is
- not: a red "X" in the middle, marks the point where four tatami meet.
- Because of this rule, certain even-sized rooms cannot be covered with
- tatami: we call them tatami-free rooms.
- Further, we define T(s) as the number of tatami-free rooms of size s.
- The smallest tatami-free room has size s = 70 and dimensions 7×10.
- All the other rooms of size s = 70 can be covered with tatami; they are:
- 1×70, 2×35 and 5×14.
- Hence, T(70) = 1.
- Similarly, we can verify that T(1320) = 5 because there are exactly 5
- tatami-free rooms of size s = 1320:
- 20×66, 22×60, 24×55, 30×44 and 33×40.
- In fact, s = 1320 is the smallest room-size s for which T(s) = 5.
- Find the smallest room-size s for which T(s) = 200.
- p_256_tatami3.gif
- Answer: 85765680
- Problem 257
- ===========
- Given is an integer sided triangle ABC with sides a ≤ b ≤ c. (AB = c, BC =
- a and AC = b).
- The angular bisectors of the triangle intersect the sides at points E, F
- and G (see picture below).
- The segments EF, EG and FG partition the triangle ABC into four smaller
- triangles: AEG, BFE, CGF and EFG.
- It can be proven that for each of these four triangles the ratio
- area(ABC)/area(subtriangle) is rational.
- However, there exist triangles for which some or all of these ratios are
- integral.
- How many triangles ABC with perimeter≤100,000,000 exist so that the ratio
- area(ABC)/area(AEG) is integral?
- p_257_bisector.gif
- Answer: 139012411
- Problem 258
- ===========
- A sequence is defined as:
- • g[k] = 1, for 0 ≤ k ≤ 1999
- • g[k] = g[k-2000] + g[k-1999], for k ≥ 2000.
- Find g[k] mod 20092010 for k = 10^18.
- Answer: 12747994
- Problem 259
- ===========
- A positive integer will be called reachable if it can result from an
- arithmetic expression obeying the following rules:
- • Uses the digits 1 through 9, in that order and exactly once each.
- • Any successive digits can be concatenated (for example, using the
- digits 2, 3 and 4 we obtain the number 234).
- • Only the four usual binary arithmetic operations (addition,
- subtraction, multiplication and division) are allowed.
- • Each operation can be used any number of times, or not at all.
- • Unary minus is not allowed.
- • Any number of (possibly nested) parentheses may be used to define the
- order of operations.
- For example, 42 is reachable, since (1/23) * ((4*5)-6) * (78-9) = 42.
- What is the sum of all positive reachable integers?
- Answer: 771828a57c269d873335c9091af78f76
- Problem 260
- ===========
- A game is played with three piles of stones and two players.
- At her turn, a player removes one or more stones from the piles. However,
- if she takes stones from more than one pile, she must remove the same
- number of stones from each of the selected piles.
- In other words, the player chooses some N>0 and removes:
- • N stones from any single pile; or
- • N stones from each of any two piles (2N total); or
- • N stones from each of the three piles (3N total).
- The player taking the last stone(s) wins the game.
- A winning configuration is one where the first player can force a win.
- For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations
- because the first player can immediately remove all stones.
- A losing configuration is one where the second player can force a win, no
- matter what the first player does.
- For example, (0,1,2) and (1,3,3) are losing configurations: any legal move
- leaves a winning configuration for the second player.
- Consider all losing configurations (x[i],y[i],z[i]) where x[i] ≤ y[i] ≤
- z[i] ≤ 100.
- We can verify that Σ(x[i]+y[i]+z[i]) = 173895 for these.
- Find Σ(x[i]+y[i]+z[i]) where (x[i],y[i],z[i]) ranges over the losing
- configurations
- with x[i] ≤ y[i] ≤ z[i] ≤ 1000.
- Answer: 167542057
- Problem 261
- ===========
- Let us call a positive integer k a square-pivot, if there is a pair of
- integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive
- squares up to k equals the sum of the m consecutive squares from (n+1) on:
- (k-m)^2 + ... + k^2 = (n+1)^2 + ... + (n+m)^2.
- Some small square-pivots are
- • 4: 3^2 + 4^2 = 5^2
- • 21: 20^2 + 21^2 = 29^2
- • 24: 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
- • 110: 108^2 + 109^2 + 110^2 = 133^2 + 134^2
- Find the sum of all distinct square-pivots ≤ 10^10.
- Answer: d45ddf64010ed143228a6a6b84837de9
- Problem 262
- ===========
- The following equation represents the continuous topography of a
- mountainous region, giving the elevation h at any point (x,y):
- A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving
- the area given by 0 ≤ x, y ≤ 1600.
- Because of the intervening mountains, it first rises straight up to a
- point A', having elevation f. Then, while remaining at the same elevation
- f, it flies around any obstacles until it arrives at a point B' directly
- above B.
- First, determine f[min] which is the minimum constant elevation allowing
- such a trip from A to B, while remaining in the specified area.
- Then, find the length of the shortest path between A' and B', while flying
- at that constant elevation f[min].
- Give that length as your answer, rounded to three decimal places.
- Note: For convenience, the elevation function shown above is repeated
- below, in a form suitable for most programming languages:
- h=( 5000-0.005*(x*x+y*y+x*y)+12.5*(x+y) ) * exp(
- -abs(0.000001*(x*x+y*y)-0.0015*(x+y)+0.7) )
- p_262_formula1.gif
- Answer: a5921e175a44d31e7f82f7f9a61a36af
- Problem 263
- ===========
- Consider the number 6. The divisors of 6 are: 1,2,3 and 6.
- Every number from 1 up to and including 6 can be written as a sum of
- distinct divisors of 6:
- 1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6.
- A number n is called a practical number if every number from 1 up to and
- including n can be expressed as a sum of distinct divisors of n.
- A pair of consecutive prime numbers with a difference of six is called a
- sexy pair (since "sex" is the Latin word for "six"). The first sexy pair
- is (23, 29).
- We may occasionally find a triple-pair, which means three consecutive sexy
- prime pairs, such that the second member of each pair is the first member
- of the next pair.
- We shall call a number n such that :
- • (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and
- • the numbers n-8, n-4, n, n+4 and n+8 are all practical,
- an engineers’ paradise.
- Find the sum of the first four engineers’ paradises.
- Answer: 8fe3eb7196c69a080740e076cff9b4a1
- Problem 264
- ===========
- Consider all the triangles having:
- • All their vertices on lattice points.
- • Circumcentre at the origin O.
- • Orthocentre at the point H(5, 0).
- There are nine such triangles having a perimeter ≤ 50.
- Listed and shown in ascending order of their perimeter, they are:
- A(-4, 3), B(5, 0), C(4, -3)
- A(4, 3), B(5, 0), C(-4, -3)
- A(-3, 4), B(5, 0), C(3, -4)
- A(3, 4), B(5, 0), C(-3, -4)
- A(0, 5), B(5, 0), C(0, -5)
- A(1, 8), B(8, -1), C(-4, -7)
- A(8, 1), B(1, -8), C(-4, 7)
- A(2, 9), B(9, -2), C(-6, -7)
- A(9, 2), B(2, -9), C(-6, 7)
- The sum of their perimeters, rounded to four decimal places, is 291.0089.
- Find all such triangles with a perimeter ≤ 10^5.
- Enter as your answer the sum of their perimeters rounded to four decimal
- places.
- p_264_TriangleCentres.gif
- Answer: 287514a045a38be0a75a1786694c77ee
- Problem 265
- ===========
- 2^N binary digits can be placed in a circle so that all the N-digit
- clockwise subsequences are distinct.
- For N=3, two such circular arrangements are possible, ignoring rotations:
- For the first arrangement, the 3-digit subsequences, in clockwise order,
- are:
- 000, 001, 010, 101, 011, 111, 110 and 100.
- Each circular arrangement can be encoded as a number by concatenating the
- binary digits starting with the subsequence of all zeros as the most
- significant bits and proceeding clockwise. The two arrangements for N=3
- are thus represented as 23 and 29:
- 00010111 [2] = 23
- 00011101 [2] = 29
- Calling S(N) the sum of the unique numeric representations, we can see
- that S(3) = 23 + 29 = 52.
- Find S(5).
- p_265_BinaryCircles.gif
- Answer: c25cebbc8dce4bdcf96cb395a11afcc3
- Problem 266
- ===========
- The divisors of 12 are: 1,2,3,4,6 and 12.
- The largest divisor of 12 that does not exceed the square root of 12 is 3.
- We shall call the largest divisor of an integer n that does not exceed the
- square root of n the pseudo square root (PSR) of n.
- It can be seen that PSR(3102)=47.
- Let p be the product of the primes below 190.
- Find PSR(p) mod 10^16.
- Answer: 32da1d501e539ab509f104e2db68d57a
- Problem 267
- ===========
- You are given a unique investment opportunity.
- Starting with £1 of capital, you can choose a fixed proportion, f, of your
- capital to bet on a fair coin toss repeatedly for 1000 tosses.
- Your return is double your bet for heads and you lose your bet for tails.
- For example, if f = 1/4, for the first toss you bet £0.25, and if heads
- comes up you win £0.5 and so then have £1.5. You then bet £0.375 and if
- the second toss is tails, you have £1.125.
- Choosing f to maximize your chances of having at least £1,000,000,000
- after 1,000 flips, what is the chance that you become a billionaire?
- All computations are assumed to be exact (no rounding), but give your
- answer rounded to 12 digits behind the decimal point in the form
- 0.abcdefghijkl.
- Answer: b8dd3306c2c64eacb0ac36b414892edb
- Problem 268
- ===========
- It can be verified that there are 23 positive integers less than 1000 that
- are divisible by at least four distinct primes less than 100.
- Find how many positive integers less than 10^16 are divisible by at least
- four distinct primes less than 100.
- Answer: 6f84b20c10311cb24a824416a3c3e0a4
- Problem 269
- ===========
- A root or zero of a polynomial P(x) is a solution to the equation P(x) =
- 0.
- Define P[n] as the polynomial whose coefficients are the digits of n.
- For example, P[5703](x) = 5x^3 + 7x^2 + 3.
- We can see that:
- • P[n](0) is the last digit of n,
- • P[n](1) is the sum of the digits of n,
- • P[n](10) is n itself.
- Define Z(k) as the number of positive integers, n, not exceeding k for
- which the polynomial P[n] has at least one integer root.
- It can be verified that Z(100 000) is 14696.
- What is Z(10^16)?
- Answer: f7ba868cb52a9b9c7e58b1b92e230be8
- Problem 270
- ===========
- A square piece of paper with integer dimensions N×N is placed with a
- corner at the origin and two of its sides along the x- and y-axes. Then,
- we cut it up respecting the following rules:
- • We only make straight cuts between two points lying on different sides
- of the square, and having integer coordinates.
- • Two cuts cannot cross, but several cuts can meet at the same border
- point.
- • Proceed until no more legal cuts can be made.
- Counting any reflections or rotations as distinct, we call C(N) the number
- of ways to cut an N×N square. For example, C(1) = 2 and C(2) = 30 (shown
- below).
- What is C(30) mod 10^8 ?
- p_270_CutSquare.gif
- Answer: 82282080
- Problem 271
- ===========
- For a positive number n, define S(n) as the sum of the integers x, for
- which 1<x<n and
- x^3≡1 mod n.
- When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53,
- 74, 79, 81.
- Thus, S(91)=9+16+22+29+53+74+79+81=363.
- Find S(13082761331670030).
- Answer: c4157aab542bd0dfa465c890e1286cc5
- Problem 272
- ===========
- For a positive number n, define C(n) as the number of the integers x, for
- which 1<x<n and
- x^3≡1 mod n.
- When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53,
- 74, 79, 81.
- Thus, C(91)=8.
- Find the sum of the positive numbers n≤10^11 for which C(n)=242.
- Answer: d84d2020055b3e8867dc359e739e0312
- Problem 273
- ===========
- Consider equations of the form: a^2 + b^2 = N, 0 ≤ a ≤ b, a, b and N
- integer.
- For N=65 there are two solutions:
- a=1, b=8 and a=4, b=7.
- We call S(N) the sum of the values of a of all solutions of a^2 + b^2 = N,
- 0 ≤ a ≤ b, a, b and N integer.
- Thus S(65) = 1 + 4 = 5.
- Find ∑S(N), for all squarefree N only divisible by primes of the form 4k+1
- with 4k+1 < 150.
- Answer: 2b03731e58e9d60e559ee5fdce4f0d14
- Problem 274
- ===========
- For each integer p > 1 coprime to 10 there is a positive divisibility
- multiplier m < p which preserves divisibility by p for the following
- function on any positive integer, n:
- f(n) = (all but the last digit of n) + (the last digit of n) * m
- That is, if m is the divisibility multiplier for p, then f(n) is divisible
- by p if and only if n is divisible by p.
- (When n is much larger than p, f(n) will be less than n and repeated
- application of f provides a multiplicative divisibility test for p.)
- For example, the divisibility multiplier for 113 is 34.
- f(76275) = 7627 + 5 * 34 = 7797 : 76275 and 7797 are both divisible by 113
- f(12345) = 1234 + 5 * 34 = 1404 : 12345 and 1404 are both not divisible by
- 113
- The sum of the divisibility multipliers for the primes that are coprime to
- 10 and less than 1000 is 39517. What is the sum of the divisibility
- multipliers for the primes that are coprime to 10 and less than 10^7?
- Answer: ffd68ca67b9c3ea2653d375051e70288
- Problem 275
- ===========
- Let us define a balanced sculpture of order n as follows:
- • A polyomino made up of n+1 tiles known as the blocks (n tiles)
- and the plinth (remaining tile);
- • the plinth has its centre at position (x = 0, y = 0);
- • the blocks have y-coordinates greater than zero (so the plinth is the
- unique lowest tile);
- • the centre of mass of all the blocks, combined, has x-coordinate equal
- to zero.
- When counting the sculptures, any arrangements which are simply
- reflections about the y-axis, are not counted as distinct. For example,
- the 18 balanced sculptures of order 6 are shown below; note that each pair
- of mirror images (about the y-axis) is counted as one sculpture:
- There are 964 balanced sculptures of order 10 and 360505 of order 15.
- How many balanced sculptures are there of order 18?
- p_275_sculptures2.gif
- Answer: 15030564
- Problem 276
- ===========
- Consider the triangles with integer sides a, b and c with a ≤ b ≤ c.
- An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1.
- How many primitive integer sided triangles exist with a perimeter not
- exceeding 10 000 000?
- Answer: 29ae64e74ebfdf459dac56786e95c5d5
- Problem 277
- ===========
- A modified Collatz sequence of integers is obtained from a starting value
- a[1] in the following way:
- a[n+1] = a[n]/3 if a[n] is divisible by 3. We shall denote this as a large
- downward step, "D".
- a[n+1] = (4a[n] + 2)/3 if a[n] divided by 3 gives a remainder of 1. We
- shall denote this as an upward step, "U".
- a[n+1] = (2a[n] - 1)/3 if a[n] divided by 3 gives a remainder of 2. We
- shall denote this as a small downward step, "d".
- The sequence terminates when some a[n] = 1.
- Given any integer, we can list out the sequence of steps.
- For instance if a[1]=231, then the sequence
- {a[n]}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps
- "DdDddUUdDD".
- Of course, there are other sequences that begin with that same sequence
- "DdDddUUdDD....".
- For instance, if a[1]=1004064, then the sequence is
- DdDddUUdDDDdUDUUUdDdUUDDDUdDD.
- In fact, 1004064 is the smallest possible a[1] > 10^6 that begins with the
- sequence DdDddUUdDD.
- What is the smallest a[1] > 10^15 that begins with the sequence
- "UDDDUdddDDUDDddDdDddDDUDDdUUDd"?
- Answer: 9508afff135320c18d82c93a8b70cd11
- Problem 278
- ===========
- Given the values of integers 1 < a[1] < a[2] <... < a[n], consider the
- linear combination
- q[1]a[1] + q[2]a[2] + ... + q[n]a[n] = b, using only integer values q[k] ≥
- 0.
- Note that for a given set of a[k], it may be that not all values of b are
- possible.
- For instance, if a[1] = 5 and a[2] = 7, there are no q[1] ≥ 0 and q[2] ≥ 0
- such that b could be
- 1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23.
- In fact, 23 is the largest impossible value of b for a[1] = 5 and a[2] =
- 7.
- We therefore call f(5, 7) = 23.
- Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195.
- Find ∑ f(p*q,p*r,q*r), where p, q and r are prime numbers and p < q < r <
- 5000.
- Answer: 7e680606b5e9890a19894dbdbbbd102a
- Problem 279
- ===========
- How many triangles are there with integral sides, at least one integral
- angle (measured in degrees), and a perimeter that does not exceed 10^8?
- Answer: 1f51455a8180fdeeb21285dfb6cba45f
- Problem 280
- ===========
- A laborious ant walks randomly on a 5x5 grid. The walk starts from the
- central square. At each step, the ant moves to an adjacent square at
- random, without leaving the grid; thus there are 2, 3 or 4 possible moves
- at each step depending on the ant's position.
- At the start of the walk, a seed is placed on each square of the lower
- row. When the ant isn't carrying a seed and reaches a square of the lower
- row containing a seed, it will start to carry the seed. The ant will drop
- the seed on the first empty square of the upper row it eventually reaches.
- What's the expected number of steps until all seeds have been dropped in
- the top row?
- Give your answer rounded to 6 decimal places.
- Answer: 27f07f04d1908e5ce4fa6eac09881cc2
- Problem 281
- ===========
- You are given a pizza (perfect circle) that has been cut into m·n equal
- pieces and you want to have exactly one topping on each slice.
- Let f(m,n) denote the number of ways you can have toppings on the pizza
- with m different toppings (m ≥ 2), using each topping on exactly n slices
- (n ≥ 1).
- Reflections are considered distinct, rotations are not.
- Thus, for instance, f(2,1) = 1, f(2,2) = f(3,1) = 2 and f(3,2) = 16.
- f(3,2) is shown below:
- Find the sum of all f(m,n) such that f(m,n) ≤ 10^15.
- p_281_pizza.gif
- Answer: ceee6ced9d64aad844310c8ce2aae2b7
- Problem 282
- ===========
- For non-negative integers m, n, the Ackermann function A(m, n) is defined
- as follows:
- For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125.
- Find A(n, n) and give your answer mod 14^8.
- p_282_formula.gif
- Answer: a1cc665e127af4e907e13087ee777bd5
- Problem 283
- ===========
- Consider the triangle with sides 6, 8 and 10. It can be seen that the
- perimeter and the area are both equal to 24. So the area/perimeter ratio
- is equal to 1.
- Consider also the triangle with sides 13, 14 and 15. The perimeter equals
- 42 while the area is equal to 84. So for this triangle the area/perimeter
- ratio is equal to 2.
- Find the sum of the perimeters of all integer sided triangles for which
- the area/perimeter ratios are equal to positive integers not exceeding
- 1000.
- Answer: 08afda4bc05c8f3ef71c9ffea1ddc0c8
- Problem 284
- ===========
- The 3-digit number 376 in the decimal numbering system is an example of
- numbers with the special property that its square ends with the same
- digits: 376^2 = 141376. Let's call a number with this property a steady
- square.
- Steady squares can also be observed in other numbering systems. In the
- base 14 numbering system, the 3-digit number c37 is also a steady square:
- c37^2 = aa0c37, and the sum of its digits is c+3+7=18 in the same
- numbering system. The letters a, b, c and d are used for the 10, 11, 12
- and 13 digits respectively, in a manner similar to the hexadecimal
- numbering system.
- For 1 ≤ n ≤ 9, the sum of the digits of all the n-digit steady squares in
- the base 14 numbering system is 2d8 (582 decimal). Steady squares with
- leading 0's are not allowed.
- Find the sum of the digits of all the n-digit steady squares in the base
- 14 numbering system for
- 1 ≤ n ≤ 10000 (decimal) and give your answer in the base 14 system using
- lower case letters where necessary.
- Answer: aff724582e583649876f518f9b340a69
- Problem 285
- ===========
- Albert chooses a positive integer k, then two real numbers a, b are
- randomly chosen in the interval [0,1] with uniform distribution.
- The square root of the sum (k·a+1)^2 + (k·b+1)^2 is then computed and
- rounded to the nearest integer. If the result is equal to k, he scores k
- points; otherwise he scores nothing.
- For example, if k = 6, a = 0.2 and b = 0.85, then
- (k·a+1)^2 + (k·b+1)^2 = 42.05.
- The square root of 42.05 is 6.484... and when rounded to the nearest
- integer, it becomes 6.
- This is equal to k, so he scores 6 points.
- It can be shown that if he plays 10 turns with k = 1, k = 2, ..., k = 10,
- the expected value of his total score, rounded to five decimal places, is
- 10.20914.
- If he plays 10^5 turns with k = 1, k = 2, k = 3, ..., k = 10^5, what is
- the expected value of his total score, rounded to five decimal places?
- Answer: bbae95d0ce2999cae57782c3746aecb6
- Problem 286
- ===========
- Barbara is a mathematician and a basketball player. She has found that the
- probability of scoring a point when shooting from a distance x is exactly
- (1 - ^x/[q]), where q is a real constant greater than 50.
- During each practice run, she takes shots from distances x = 1, x = 2,
- ..., x = 50 and, according to her records, she has precisely a 2 % chance
- to score a total of exactly 20 points.
- Find q and give your answer rounded to 10 decimal places.
- Answer: cc5a1ef0deabf698733bcef4f1149498
- Problem 287
- ===========
- The quadtree encoding allows us to describe a 2^N×2^N black and white
- image as a sequence of bits (0 and 1). Those sequences are to be read from
- left to right like this:
- • the first bit deals with the complete 2^N×2^N region;
- • "0" denotes a split:
- the current 2^n×2^n region is divided into 4 sub-regions of dimension
- 2^n-1×2^n-1,
- the next bits contains the description of the top left, top right,
- bottom left and bottom right sub-regions - in that order;
- • "10" indicates that the current region contains only black pixels;
- • "11" indicates that the current region contains only white pixels.
- Consider the following 4×4 image (colored marks denote places where a
- split can occur):
- This image can be described by several sequences, for example
- :"001010101001011111011010101010", of length 30, or
- "0100101111101110", of length 16, which is the minimal sequence for this
- image.
- For a positive integer N, define D[N] as the 2^N×2^N image with the
- following coloring scheme:
- • the pixel with coordinates x = 0, y = 0 corresponds to the bottom left
- pixel,
- • if (x - 2^N-1)^2 + (y - 2^N-1)^2 ≤ 2^2N-2 then the pixel is black,
- • otherwise the pixel is white.
- What is the length of the minimal sequence describing D[24] ?
- p_287_quadtree.gif
- Answer: 6c2beec8a6c0bc788d5e45c317b0d7ca
- Problem 288
- ===========
- For any prime p the number N(p,q) is defined byN(p,q) = ∑[n=0 to q]
- T[n]*p^n
- with T[n] generated by the following random number generator:
- S[0] = 290797
- S[n+1] = S[n]^2 mod 50515093
- T[n] = S[n] mod p
- Let Nfac(p,q) be the factorial of N(p,q).
- Let NF(p,q) be the number of factors p in Nfac(p,q).
- You are given that NF(3,10000) mod 3^20=624955285.
- Find NF(61,10^7) mod 61^10
- Answer: 192bf4aa33ea85e922d583f60fe99955
- Problem 289
- ===========
- Let C(x,y) be a circle passing through the points (x, y), (x, y+1),
- (x+1, y) and (x+1, y+1).
- For positive integers m and n, let E(m,n) be a configuration which
- consists of the m·n circles:
- { C(x,y): 0 ≤ x < m, 0 ≤ y < n, x and y are integers }
- An Eulerian cycle on E(m,n) is a closed path that passes through each arc
- exactly once.
- Many such paths are possible on E(m,n), but we are only interested in
- those which are not self-crossing: A non-crossing path just touches itself
- at lattice points, but it never crosses itself.
- The image below shows E(3,3) and an example of an Eulerian non-crossing
- path.
- Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n).
- For example, L(1,2) = 2, L(2,2) = 37 and L(3,3) = 104290.
- Find L(6,10) mod 10^10.
- p_289_euler.gif
- Answer: 9fa32696df356b3d41faa7dd278c88a9
- Problem 290
- ===========
- How many integers 0 ≤ n < 10^18 have the property that the sum of the
- digits of n equals the sum of digits of 137n?
- Answer: 8246684fec8ece9f0ee3c9898c8c9d6a
- Problem 291
- ===========
- A prime number p is called a Panaitopol prime if for some positive
- integers
- x and y.
- Find how many Panaitopol primes are less than 5×10^15.
- p_291_formula.gif
- Answer: 4037526
- Problem 292
- ===========
- We shall define a pythagorean polygon to be a convex polygon with the
- following properties:
- • there are at least three vertices,
- • no three vertices are aligned,
- • each vertex has integer coordinates,
- • each edge has integer length.
- For a given integer n, define P(n) as the number of distinct pythagorean
- polygons for which the perimeter is ≤ n.
- Pythagorean polygons should be considered distinct as long as none is a
- translation of another.
- You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045.
- Find P(120).
- Answer: 27f50f02ef10f170379b144435e0144b
- Problem 293
- ===========
- An even positive integer N will be called admissible, if it is a power of
- 2 or its distinct prime factors are consecutive primes.
- The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48.
- If N is admissible, the smallest integer M > 1 such that N+M is prime,
- will be called the pseudo-Fortunate number for N.
- For example, N=630 is admissible since it is even and its distinct prime
- factors are the consecutive primes 2,3,5 and 7.
- The next prime number after 631 is 641; hence, the pseudo-Fortunate number
- for 630 is M=11.
- It can also be seen that the pseudo-Fortunate number for 16 is 3.
- Find the sum of all distinct pseudo-Fortunate numbers for admissible
- numbers N less than 10^9.
- Answer: 2209
- Problem 294
- ===========
- For a positive integer k, define d(k) as the sum of the digits of k in its
- usual decimal representation.Thus d(42) = 4+2 = 6.
- For a positive integer n, define S(n) as the number of positive integers k
- < 10^n with the following properties :
- • k is divisible by 23 and
- • d(k) = 23.
- You are given that S(9) = 263626 and S(42) = 6377168878570056.
- Find S(11^12) and give your answer mod 10^9.
- Answer: aefe049404a284c7d27fab3887c6c4a2
- Problem 295
- ===========
- We call the convex area enclosed by two circles a lenticular hole if:
- • The centres of both circles are on lattice points.
- • The two circles intersect at two distinct lattice points.
- • The interior of the convex area enclosed by both circles does not
- contain any lattice points.
- Consider the circles:
- C[0]: x^2+y^2=25
- C[1]: (x+4)^2+(y-4)^2=1
- C[2]: (x-12)^2+(y-4)^2=65
- The circles C[0], C[1] and C[2] are drawn in the picture below.
- C[0] and C[1] form a lenticular hole, as well as C[0] and C[2].
- We call an ordered pair of positive real numbers (r[1], r[2]) a lenticular
- pair if there exist two circles with radii r[1] and r[2] that form a
- lenticular hole.We can verify that (1, 5) and (5, √65) are the lenticular
- pairs of the example above.
- Let L(N) be the number of distinct lenticular pairs (r[1], r[2]) for which
- 0 < r[1] ≤ r[2] ≤ N.
- We can verify that L(10) = 30 and L(100) = 3442.
- Find L(100 000).
- Answer: 5beaace6425205fe879116ee07dae961
- Problem 296
- ===========
- Given is an integer sided triangle ABC with BC ≤ AC ≤ AB.
- k is the angular bisector of angle ACB.
- m is the tangent at C to the circumscribed circle of ABC.
- n is a line parallel to m through B.
- The intersection of n and k is called E.
- How many triangles ABC with a perimeter not exceeding 100 000 exist such
- that BE has integral length?
- Answer: 45986a4405b2dd6c163516319e0c4a1b
- Problem 297
- ===========
- Each new term in the Fibonacci sequence is generated by adding the
- previous two terms.
- Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21,
- 34, 55, 89.
- Every positive integer can be uniquely written as a sum of nonconsecutive
- terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89.
- Such a sum is called the Zeckendorf representation of the number.
- For any integer n>0, let z(n) be the number of terms in the Zeckendorf
- representation of n.
- Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc.
- Also, for 0<n<10^6, ∑ z(n) = 7894453.
- Find ∑ z(n) for 0<n<10^17.
- Answer: d3fd75f5447698748a826562750a1986
- Problem 298
- ===========
- Larry and Robin play a memory game involving of a sequence of random
- numbers between 1 and 10, inclusive, that are called out one at a time.
- Each player can remember up to 5 previous numbers. When the called number
- is in a player's memory, that player is awarded a point. If it's not, the
- player adds the called number to his memory, removing another number if
- his memory is full.
- Both players start with empty memories. Both players always add new missed
- numbers to their memory but use a different strategy in deciding which
- number to remove:
- Larry's strategy is to remove the number that hasn't been called in the
- longest time.
- Robin's strategy is to remove the number that's been in the memory the
- longest time.
- Example game:
- Turn Called Larry's Larry's Robin's Robin's
- number memory score memory score
- 1 1 1 0 1 0
- 2 2 1,2 0 1,2 0
- 3 4 1,2,4 0 1,2,4 0
- 4 6 1,2,4,6 0 1,2,4,6 0
- 5 1 1,2,4,6 1 1,2,4,6 1
- 6 8 1,2,4,6,8 1 1,2,4,6,8 1
- 7 10 1,4,6,8,10 1 2,4,6,8,10 1
- 8 2 1,2,6,8,10 1 2,4,6,8,10 2
- 9 4 1,2,4,8,10 1 2,4,6,8,10 3
- 10 1 1,2,4,8,10 2 1,4,6,8,10 3
- Denoting Larry's score by L and Robin's score by R, what is the expected
- value of |L-R| after 50 turns? Give your answer rounded to eight decimal
- places using the format x.xxxxxxxx .
- Answer: d078fd564995aa2a813a29f44ad79611
- Problem 299
- ===========
- Four points with integer coordinates are selected:
- A(a, 0), B(b, 0), C(0, c) and D(0, d), with 0 < a < b and 0 < c < d.
- Point P, also with integer coordinates, is chosen on the line AC so that
- the three triangles ABP, CDP and BDP are all similar.
- It is easy to prove that the three triangles can be similar, only if a=c.
- So, given that a=c, we are looking for triplets (a,b,d) such that at least
- one point P (with integer coordinates) exists on AC, making the three
- triangles ABP, CDP and BDP all similar.
- For example, if (a,b,d)=(2,3,4), it can be easily verified that point
- P(1,1) satisfies the above condition. Note that the triplets (2,3,4) and
- (2,4,3) are considered as distinct, although point P(1,1) is common for
- both.
- If b+d < 100, there are 92 distinct triplets (a,b,d) such that point P
- exists.
- If b+d < 100 000, there are 320471 distinct triplets (a,b,d) such that
- point P exists.
- If b+d < 100 000 000, how many distinct triplets (a,b,d) are there such
- that point P exists?
- p_299_ThreeSimTri.gif
- Answer: fb8f093361a6db56c8a1d1661ab229cd
- Problem 300
- ===========
- In a very simplified form, we can consider proteins as strings consisting
- of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH.
- For this problem, the orientation of a protein is important; e.g. HPP is
- considered distinct from PPH. Thus, there are 2^n distinct proteins
- consisting of n elements.
- When one encounters these strings in nature, they are always folded in
- such a way that the number of H-H contact points is as large as possible,
- since this is energetically advantageous.
- As a result, the H-elements tend to accumulate in the inner part, with the
- P-elements on the outside.
- Natural proteins are folded in three dimensions of course, but we will
- only consider protein folding in two dimensions.
- The figure below shows two possible ways that our example protein could be
- folded (H-H contact points are shown with red dots).
- The folding on the left has only six H-H contact points, thus it would
- never occur naturally.
- On the other hand, the folding on the right has nine H-H contact points,
- which is optimal for this string.
- Assuming that H and P elements are equally likely to occur in any position
- along the string, the average number of H-H contact points in an optimal
- folding of a random protein string of length 8 turns out to be
- 850 / 2^8=3.3203125.
- What is the average number of H-H contact points in an optimal folding of
- a random protein string of length 15?
- Give your answer using as many decimal places as necessary for an exact
- result.
- p_300_protein.gif
- Answer: 5a0d6315bc18279c46a1fb8cbd2f16b5
- Problem 301
- ===========
- Nim is a game played with heaps of stones, where two players take it in
- turn to remove any number of stones from any heap until no stones remain.
- We'll consider the three-heap normal-play version of Nim, which works as
- follows:
- - At the start of the game there are three heaps of stones.
- - On his turn the player removes any positive number of stones from any
- single heap.
- - The first player unable to move (because no stones remain) loses.
- If (n[1],n[2],n[3]) indicates a Nim position consisting of heaps of size
- n[1], n[2] and n[3] then there is a simple function X(n[1],n[2],n[3]) —
- that you may look up or attempt to deduce for yourself — that returns:
- • zero if, with perfect strategy, the player about to move will
- eventually lose; or
- • non-zero if, with perfect strategy, the player about to move will
- eventually win.
- For example X(1,2,3) = 0 because, no matter what the current player does,
- his opponent can respond with a move that leaves two heaps of equal size,
- at which point every move by the current player can be mirrored by his
- opponent until no stones remain; so the current player loses. To
- illustrate:
- - current player moves to (1,2,1)
- - opponent moves to (1,0,1)
- - current player moves to (0,0,1)
- - opponent moves to (0,0,0), and so wins.
- For how many positive integers n ≤ 2^30 does X(n,2n,3n) = 0 ?
- Answer: 2178309
- Problem 302
- ===========
- A positive integer n is powerful if p^2 is a divisor of n for every prime
- factor p in n.
- A positive integer n is a perfect power if n can be expressed as a power
- of another positive integer.
- A positive integer n is an Achilles number if n is powerful but not a
- perfect power. For example, 864 and 1800 are Achilles numbers: 864 =
- 2^5·3^3 and 1800 = 2^3·3^2·5^2.
- We shall call a positive integer S a Strong Achilles number if both S and
- φ(S) are Achilles numbers.^1
- For example, 864 is a Strong Achilles number: φ(864) = 288 = 2^5·3^2.
- However, 1800 isn't a Strong Achilles number because: φ(1800) = 480 =
- 2^5·3^1·5^1.
- There are 7 Strong Achilles numbers below 10^4 and 656 below 10^8.
- How many Strong Achilles numbers are there below 10^18?
- ^1 φ denotes Euler's totient function.
- Answer: 1170060
- Problem 303
- ===========
- For a positive integer n, define f(n) as the least positive multiple of n
- that, written in base 10, uses only digits ≤ 2.
- Thus f(2)=2, f(3)=12, f(7)=21, f(42)=210, f(89)=1121222.
- Also, .
- Find .
- Answer: b904a0b3d922e628a828e744ee7d3a60
- Problem 304
- ===========
- For any positive integer n the function next_prime(n) returns the smallest
- prime p
- such that p>n.
- The sequence a(n) is defined by:
- a(1)=next_prime(10^14) and a(n)=next_prime(a(n-1)) for n>1.
- The fibonacci sequence f(n) is defined by:f(0)=0, f(1)=1 and
- f(n)=f(n-1)+f(n-2) for n>1.
- The sequence b(n) is defined as f(a(n)).
- Find ∑b(n) for 1≤n≤100 000. Give your answer mod 1234567891011.
- Answer: 499427a3e4bf9ad34a6df3056604b4c1
- Problem 305
- ===========
- Let's call S the (infinite) string that is made by concatenating the
- consecutive positive integers (starting from 1) written down in base 10.
- Thus, S = 1234567891011121314151617181920212223242...
- It's easy to see that any number will show up an infinite number of times
- in S.
- Let's call f(n) the starting position of the n^th occurrence of n in S.
- For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.
- Find ∑f(3^k) for 1≤k≤13.
- Answer: 9def85298f598867d361e4afca8cdd96
- Problem 306
- ===========
- The following game is a classic example of Combinatorial Game Theory:
- Two players start with a strip of n white squares and they take alternate
- turns.
- On each turn, a player picks two contiguous white squares and paints them
- black.
- The first player who cannot make a move loses.
- • If n = 1, there are no valid moves, so the first player loses
- automatically.
- • If n = 2, there is only one valid move, after which the second player
- loses.
- • If n = 3, there are two valid moves, but both leave a situation where
- the second player loses.
- • If n = 4, there are three valid moves for the first player; she can
- win the game by painting the two middle squares.
- • If n = 5, there are four valid moves for the first player (shown below
- in red); but no matter what she does, the second player (blue) wins.
- So, for 1 ≤ n ≤ 5, there are 3 values of n for which the first player can
- force a win.
- Similarly, for 1 ≤ n ≤ 50, there are 40 values of n for which the first
- player can force a win.
- For 1 ≤ n ≤ 1 000 000, how many values of n are there for which the first
- player can force a win?
- p_306_pstrip.gif
- Answer: 852938
- Problem 307
- ===========
- k defects are randomly distributed amongst n integrated-circuit chips
- produced by a factory (any number of defects may be found on a chip and
- each defect is independent of the other defects).
- Let p(k,n) represent the probability that there is a chip with at least 3
- defects.
- For instance p(3,7) ≈ 0.0204081633.
- Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal
- places in the form 0.abcdefghij
- Answer: 0c49094fa750365e13bb20ec4a158b6d
- Problem 308
- ===========
- A program written in the programming language Fractran consists of a list
- of fractions.
- The internal state of the Fractran Virtual Machine is a positive integer,
- which is initially set to a seed value. Each iteration of a Fractran
- program multiplies the state integer by the first fraction in the list
- which will leave it an integer.
- For example, one of the Fractran programs that John Horton Conway wrote
- for prime-generation consists of the following 14 fractions:
- 17 , 78 , 19 , 23 , 29 , 77 , 95 , 77 , 1 , 11 , 13 , 15 , 1 , 55 .
- 91 85 51 38 33 29 23 19 17 13 11 2 7 1
- Starting with the seed integer 2, successive iterations of the program
- produce the sequence:
- 15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544,
- 32, 240, ...
- The powers of 2 that appear in this sequence are 2^2, 2^3, 2^5, ...
- It can be shown that all the powers of 2 in this sequence have prime
- exponents and that all the primes appear as exponents of powers of 2, in
- proper order!
- If someone uses the above Fractran program to solve Project Euler Problem
- 7 (find the 10001^st prime), how many iterations would be needed until the
- program produces 2^10001st prime ?
- Answer: 43e736dfc6478a52653814248a71771d
- Problem 309
- ===========
- In the classic "Crossing Ladders" problem, we are given the lengths x and
- y of two ladders resting on the opposite walls of a narrow, level street.
- We are also given the height h above the street where the two ladders
- cross and we are asked to find the width of the street (w).
- Here, we are only concerned with instances where all four variables are
- positive integers.
- For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56.
- In fact, for integer values x, y, h and 0 < x < y < 200, there are only
- five triplets (x,y,h) producing integer solutions for w:
- (70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175,
- 40).
- For integer values x, y, h and 0 < x < y < 1 000 000, how many triplets
- (x,y,h) produce integer solutions for w?
- p_309_ladders.gif
- Answer: 210139
- Problem 310
- ===========
- Alice and Bob play the game Nim Square.
- Nim Square is just like ordinary three-heap normal play Nim, but the
- players may only remove a square number of stones from a heap.
- The number of stones in the three heaps is represented by the ordered
- triple (a,b,c).
- If 0≤a≤b≤c≤29 then the number of losing positions for the next player is
- 1160.
- Find the number of losing positions for the next player if 0≤a≤b≤c≤100
- 000.
- Answer: 6b94f848996393eef163add4d17360c7
- Problem 311
- ===========
- ABCD is a convex, integer sided quadrilateral with 1 ≤ AB < BC < CD < AD.
- BD has integer length. O is the midpoint of BD. AO has integer length.
- We'll call ABCD a biclinic integral quadrilateral if AO = CO ≤ BO = DO.
- For example, the following quadrilateral is a biclinic integral
- quadrilateral:
- AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23.
- Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD
- that satisfy AB^2+BC^2+CD^2+AD^2 ≤ N.
- We can verify that B(10 000) = 49 and B(1 000 000) = 38239.
- Find B(10 000 000 000).
- p_311_biclinic.gif
- Answer: 36115d4f7dc07eea106d78e8431868e6
- Problem 312
- ===========
- - A Sierpiński graph of order-1 (S[1]) is an equilateral triangle.
- - S[n+1] is obtained from S[n] by positioning three copies of S[n] so that
- every pair of copies has one common corner.
- Let C(n) be the number of cycles that pass exactly once through all the
- vertices of S[n].
- For example, C(3) = 8 because eight such cycles can be drawn on S[3], as
- shown below:
- It can also be verified that :
- C(1) = C(2) = 1
- C(5) = 71328803586048
- C(10 000) mod 10^8 = 37652224
- C(10 000) mod 13^8 = 617720485
- Find C(C(C(10 000))) mod 13^8.
- p_312_sierpinskyAt.gif
- p_312_sierpinsky8t.gif
- Answer: 535113d1a81f421fe814d48205dac570
- Problem 313
- ===========
- In a sliding game a counter may slide horizontally or vertically into an
- empty space. The objective of the game is to move the red counter from the
- top left corner of a grid to the bottom right corner; the space always
- starts in the bottom right corner. For example, the following sequence of
- pictures show how the game can be completed in five moves on a 2 by 2
- grid.
- Let S(m,n) represent the minimum number of moves to complete the game on
- an m by n grid. For example, it can be verified that S(5,4) = 25.
- There are exactly 5482 grids for which S(m,n) = p^2, where p < 100 is
- prime.
- How many grids does S(m,n) = p^2, where p < 10^6 is prime?
- p_313_sliding_game_1.gif
- p_313_sliding_game_2.gif
- Answer: 2468d42fa1c7f61547ce71c9826218ea
- Problem 314
- ===========
- The moon has been opened up, and land can be obtained for free, but there
- is a catch. You have to build a wall around the land that you stake out,
- and building a wall on the moon is expensive. Every country has been
- allotted a 500 m by 500 m square area, but they will possess only that
- area which they wall in. 251001 posts have been placed in a rectangular
- grid with 1 meter spacing. The wall must be a closed series of straight
- lines, each line running from post to post.
- The bigger countries of course have built a 2000 m wall enclosing the
- entire 250 000 m^2 area. The [1]Duchy of Grand Fenwick, has a tighter
- budget, and has asked you (their Royal Programmer) to compute what shape
- would get best maximum enclosed-area/wall-length ratio.
- You have done some preliminary calculations on a sheet of paper.For a 2000
- meter wall enclosing the 250 000 m^2 area theenclosed-area/wall-length
- ratio is 125.
- Although not allowed , but to get an idea if this is anything better: if
- you place a circle inside the square area touching the four sides the area
- will be equal to π*250^2 m^2 and the perimeter will be π*500 m, so the
- enclosed-area/wall-length ratio will also be 125.
- However, if you cut off from the square four triangles with sides 75 m, 75
- m and 75√2 m the total area becomes 238750 m^2 and the perimeter becomes
- 1400+300√2 m. So this gives an enclosed-area/wall-length ratio of 130.87,
- which is significantly better.
- Find the maximum enclosed-area/wall-length ratio.
- Give your answer rounded to 8 places behind the decimal point in the form
- abc.defghijk.
- Visible links
- 1. http://en.wikipedia.org/wiki/Grand_Fenwick
- p_314_landgrab.gif
- Answer: aa457cae6f67945d50683a85a9b70230
- Problem 315
- ===========
- Sam and Max are asked to transform two digital clocks into two "digital
- root" clocks.
- A digital root clock is a digital clock that calculates digital roots step
- by step.
- When a clock is fed a number, it will show it and then it will start the
- calculation, showing all the intermediate values until it gets to the
- result.
- For example, if the clock is fed the number 137, it will show: "137" →
- "11" → "2" and then it will go black, waiting for the next number.
- Every digital number consists of some light segments: three horizontal
- (top, middle, bottom) and four vertical (top-left, top-right, bottom-left,
- bottom-right).
- Number "1" is made of vertical top-right and bottom-right, number "4" is
- made by middle horizontal and vertical top-left, top-right and
- bottom-right. Number "8" lights them all.
- The clocks consume energy only when segments are turned on/off.
- To turn on a "2" will cost 5 transitions, while a "7" will cost only 4
- transitions.
- Sam and Max built two different clocks.
- Sam's clock is fed e.g. number 137: the clock shows "137", then the panel
- is turned off, then the next number ("11") is turned on, then the panel is
- turned off again and finally the last number ("2") is turned on and, after
- some time, off.
- For the example, with number 137, Sam's clock requires:
- "137" : (2 + 5 + 4) × 2 = 22 transitions ("137" on/off).
- "11" : (2 + 2) × 2 = 8 transitions ("11" on/off).
- "2" : (5) × 2 = 10 transitions ("2" on/off).
- For a grand total of 40 transitions.
- Max's clock works differently. Instead of turning off the whole panel, it
- is smart enough to turn off only those segments that won't be needed for
- the next number.
- For number 137, Max's clock requires:
- 2 + 5 + 4 = 11 transitions ("137" on)
- "137" : 7 transitions (to turn off the segments that are not needed for
- number "11").
- 0 transitions (number "11" is already turned on correctly)
- "11" : 3 transitions (to turn off the first "1" and the bottom part of
- the second "1";
- the top part is common with number "2").
- 4 tansitions (to turn on the remaining segments in order to get a
- "2" : "2")
- 5 transitions (to turn off number "2").
- For a grand total of 30 transitions.
- Of course, Max's clock consumes less power than Sam's one.
- The two clocks are fed all the prime numbers between A = 10^7 and B =
- 2×10^7.
- Find the difference between the total number of transitions needed by
- Sam's clock and that needed by Max's one.
- p_315_clocks.gif
- Answer: 13625242
- Problem 316
- ===========
- Let p = p[1] p[2] p[3] ... be an infinite sequence of random digits,
- selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
- It can be seen that p corresponds to the real number 0.p[1] p[2] p[3] ....
- It can also be seen that choosing a random real number from the interval
- [0,1) is equivalent to choosing an infinite sequence of random digits
- selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
- For any positive integer n with d decimal digits, let k be the smallest
- index such that
- p[k, ]p[k+1], ...p[k+d-1] are the decimal digits of n, in the same order.
- Also, let g(n) be the expected value of k; it can be proven that g(n) is
- always finite and, interestingly, always an integer number.
- For example, if n = 535, then
- for p = 31415926535897...., we get k = 9
- for p = 355287143650049560000490848764084685354..., we get k = 36
- etc and we find that g(535) = 1008.
- Given that , find
- Note: represents the floor function.
- p_316_decexp1.gif
- p_316_decexp2.gif
- p_316_decexp3.gif
- Answer: 2495e8f6e9d4cdadbf0411144e7180b9
- Problem 317
- ===========
- A firecracker explodes at a height of 100 m above level ground. It breaks
- into a large number of very small fragments, which move in every
- direction; all of them have the same initial velocity of 20 m/s.
- We assume that the fragments move without air resistance, in a uniform
- gravitational field with g=9.81 m/s^2.
- Find the volume (in m^3) of the region through which the fragments move
- before reaching the ground. Give your answer rounded to four decimal
- places.
- Answer: b0e2bec93bfe598ade5d3d1141f76bdd
- Problem 318
- ===========
- Consider the real number √2+√3.
- When we calculate the even powers of √2+√3we get:
- (√2+√3)^2 = 9.898979485566356...
- (√2+√3)^4 = 97.98979485566356...
- (√2+√3)^6 = 969.998969071069263...
- (√2+√3)^8 = 9601.99989585502907...
- (√2+√3)^10 = 95049.999989479221...
- (√2+√3)^12 = 940897.9999989371855...
- (√2+√3)^14 = 9313929.99999989263...
- (√2+√3)^16 = 92198401.99999998915...
- It looks like that the number of consecutive nines at the beginning of the
- fractional part of these powers is non-decreasing.
- In fact it can be proven that the fractional part of (√2+√3)^2n approaches
- 1 for large n.
- Consider all real numbers of the form √p+√q with p and q positive integers
- and p<q, such that the fractional part of (√p+√q)^2n approaches 1 for
- large n.
- Let C(p,q,n) be the number of consecutive nines at the beginning of the
- fractional part of
- (√p+√q)^2n.
- Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
- Find ∑N(p,q) for p+q ≤ 2011.
- Answer: de358f1c4d6e30c1a4f82c8bc5cedf2d
- Problem 319
- ===========
- Let x[1], x[2],..., x[n] be a sequence of length n such that:
- • x[1] = 2
- • for all 1 < i ≤ n : x[i-1] < x[i]
- • for all i and j with 1 ≤ i, j ≤ n : (x[i]) ^ j < (x[j] + 1)^i
- There are only five such sequences of length 2, namely:{2,4}, {2,5},
- {2,6}, {2,7} and {2,8}.
- There are 293 such sequences of length 5; three examples are given below:
- {2,5,11,25,55}, {2,6,14,36,88}, {2,8,22,64,181}.
- Let t(n) denote the number of such sequences of length n.
- You are given that t(10) = 86195 and t(20) = 5227991891.
- Find t(10^10) and give your answer modulo 10^9.
- Answer: d346ab7d128ee0402820edf5fe4aed30
- Problem 320
- ===========
- Let N(i) be the smallest integer n such that n! is divisible by
- (i!)^1234567890
- Let S(u)=∑N(i) for 10 ≤ i ≤ u.
- S(1000)=614538266565663.
- Find S(1 000 000) mod 10^18.
- Answer: 8426f939c3ee410a8c4d43886ef77ccb
- Problem 321
- ===========
- A horizontal row comprising of 2n + 1 squares has n red counters placed at
- one end and n blue counters at the other end, being separated by a single
- empty square in the centre. For example, when n = 3.
- A counter can move from one square to the next (slide) or can jump over
- another counter (hop) as long as the square next to that counter is
- unoccupied.
- Let M(n) represent the minimum number of moves/actions to completely
- reverse the positions of the coloured counters; that is, move all the red
- counters to the right and all the blue counters to the left.
- It can be verified M(3) = 15, which also happens to be a triangle number.
- If we create a sequence based on the values of n for which M(n) is a
- triangle number then the first five terms would be:
- 1, 3, 10, 22, and 63, and their sum would be 99.
- Find the sum of the first forty terms of this sequence.
- p_321_swapping_counters_1.gif
- p_321_swapping_counters_2.gif
- Answer: 6d87412130312b01a999225a5fe689b1
- Problem 322
- ===========
- Let T(m, n) be the number of the binomial coefficients ^iC[n] that are
- divisible by 10 for n ≤ i < m(i, m and n are positive integers).
- You are given that T(10^9, 10^7-10) = 989697000.
- Find T(10^18, 10^12-10).
- Answer: a75af9d717fa592487fb45e7552204a8
- Problem 323
- ===========
- Let y[0], y[1], y[2],... be a sequence of random unsigned 32 bit integers
- (i.e. 0 ≤ y[i] < 2^32, every value equally likely).
- For the sequence x[i] the following recursion is given:
- • x[0] = 0 and
- • x[i] = x[i-1] | y[i-1], for i > 0. ( | is the bitwise-OR operator)
- It can be seen that eventually there will be an index N such that x[i] =
- 2^32 -1 (a bit-pattern of all ones) for all i ≥ N.
- Find the expected value of N.
- Give your answer rounded to 10 digits after the decimal point.
- Answer: c8f8a7ab17a87f1b17a1f4a86c984ea7
- Problem 324
- ===========
- Let f(n) represent the number of ways one can fill a 3×3×n tower with
- blocks of 2×1×1.
- You're allowed to rotate the blocks in any way you like; however,
- rotations, reflections etc of the tower itself are counted as distinct.
- For example (with q = 100000007) :
- f(2) = 229,
- f(4) = 117805,
- f(10) mod q = 96149360,
- f(10^3) mod q = 24806056,
- f(10^6) mod q = 30808124.
- Find f(10^10000) mod 100000007.
- Answer: 96972774
- Problem 325
- ===========
- A game is played with two piles of stones and two players. At her turn, a
- player removes a number of stones from the larger pile. The number of
- stones she removes must be a positive multiple of the number of stones in
- the smaller pile.
- E.g., let the ordered pair(6,14) describe a configuration with 6 stones in
- the smaller pile and 14 stones in the larger pile, then the first player
- can remove 6 or 12 stones from the larger pile.
- The player taking all the stones from a pile wins the game.
- A winning configuration is one where the first player can force a win. For
- example, (1,5), (2,6) and (3,12) are winning configurations because the
- first player can immediately remove all stones in the second pile.
- A losing configuration is one where the second player can force a win, no
- matter what the first player does. For example, (2,3) and (3,4) are losing
- configurations: any legal move leaves a winning configuration for the
- second player.
- Define S(N) as the sum of (x[i]+y[i]) for all losing configurations
- (x[i],y[i]), 0 < x[i] < y[i] ≤ N. We can verify that S(10) = 211 and
- S(10^4) = 230312207313.
- Find S(10^16) mod 7^10.
- Answer: 54672965
- Problem 326
- ===========
- Let a[n] be a sequence recursively defined by: .
- So the first 10 elements of a[n] are: 1,1,0,3,0,3,5,4,1,9.
- Let f(N,M) represent the number of pairs (p,q) such that:
- It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and
- (9,10).
- You are also given that f(10^4,10^3)=97158.
- Find f(10^12,10^6).
- p_326_formula1.gif
- p_326_formula2.gif
- Answer: d95dff1a5ceee0064993d98defdd603e
- Problem 327
- ===========
- A series of three rooms are connected to each other by automatic doors.
- Each door is operated by a security card. Once you enter a room the door
- automatically closes and that security card cannot be used again. A
- machine at the start will dispense an unlimited number of cards, but each
- room (including the starting room) contains scanners and if they detect
- that you are holding more than three security cards or if they detect an
- unattended security card on the floor, then all the doors will become
- permanently locked. However, each room contains a box where you may safely
- store any number of security cards for use at a later stage.
- If you simply tried to travel through the rooms one at a time then as you
- entered room 3 you would have used all three cards and would be trapped in
- that room forever!
- However, if you make use of the storage boxes, then escape is possible.
- For example, you could enter room 1 using your first card, place one card
- in the storage box, and use your third card to exit the room back to the
- start. Then after collecting three more cards from the dispensing machine
- you could use one to enter room 1 and collect the card you placed in the
- box a moment ago. You now have three cards again and will be able to
- travel through the remaining three doors. This method allows you to travel
- through all three rooms using six security cards in total.
- It is possible to travel through six rooms using a total of 123 security
- cards while carrying a maximum of 3 cards.
- Let C be the maximum number of cards which can be carried at any time.
- Let R be the number of rooms to travel through.
- Let M(C,R) be the minimum number of cards required from the dispensing
- machine to travel through R rooms carrying up to a maximum of C cards at
- any time.
- For example, M(3,6)=123 and M(4,6)=23.
- And, ΣM(C,6)=146 for 3 ≤ C ≤ 4.
- You are given that ΣM(C,10)=10382 for 3 ≤ C ≤ 10.
- Find ΣM(C,30) for 3 ≤ C ≤ 40.
- p_327_rooms_of_doom.gif
- Answer: 2cd4c0ad8a00c5be99802188ee2628fb
- Problem 328
- ===========
- We are trying to find a hidden number selected from the set of integers
- {1, 2, ..., n} by asking questions. Each number (question) we ask, has a
- cost equal to the number asked and we get one of three possible answers:
- • "Your guess is lower than the hidden number", or
- • "Yes, that's it!", or
- • "Your guess is higher than the hidden number".
- Given the value of n, an optimal strategy minimizes the total cost (i.e.
- the sum of all the questions asked) for the worst possible case. E.g.
- If n=3, the best we can do is obviously to ask the number "2". The answer
- will immediately lead us to find the hidden number (at a total cost = 2).
- If n=8, we might decide to use a "binary search" type of strategy: Our
- first question would be "4" and if the hidden number is higher than 4 we
- will need one or two additional questions.
- Let our second question be "6". If the hidden number is still higher than
- 6, we will need a third question in order to discriminate between 7 and 8.
- Thus, our third question will be "7" and the total cost for this
- worst-case scenario will be 4+6+7=17.
- We can improve considerably the worst-case cost for n=8, by asking "5" as
- our first question.
- If we are told that the hidden number is higher than 5, our second
- question will be "7", then we'll know for certain what the hidden number
- is (for a total cost of 5+7=12).
- If we are told that the hidden number is lower than 5, our second question
- will be "3" and if the hidden number is lower than 3 our third question
- will be "1", giving a total cost of 5+3+1=9.
- Since 12>9, the worst-case cost for this strategy is 12. That's better
- than what we achieved previously with the "binary search" strategy; it is
- also better than or equal to any other strategy.
- So, in fact, we have just described an optimal strategy for n=8.
- Let C(n) be the worst-case cost achieved by an optimal strategy for n, as
- described above.
- Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12.
- Similarly, C(100) = 400 and C(n) = 17575.
- Find C(n).
- p_328_sum1.gif
- p_328_sum2.gif
- Answer: 92a3220ad5b17a562c039e6e93d6df90
- Problem 329
- ===========
- Susan has a prime frog.
- Her frog is jumping around over 500 squares numbered 1 to 500.He can only
- jump one square to the left or to the right, with equal probability, and
- he cannot jump outside the range [1;500].
- (if it lands at either end, it automatically jumps to the only available
- square on the next move.)
- When he is on a square with a prime number on it, he croaks 'P' (PRIME)
- with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before
- jumping to the next square.
- When he is on a square with a number on it that is not a prime he croaks
- 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping
- to the next square.
- Given that the frog's starting position is random with the same
- probability for every square, and given that she listens to his first 15
- croaks, what is the probability that she hears the sequence
- PPPPNNPPPNPPNPN?
- Give your answer as a fraction p/q in reduced form.
- Answer: e392a8b1b053c83e68663e08456bb392
- Problem 330
- ===========
- An infinite sequence of real numbers a(n) is defined for all integers n as
- follows:
- For example,
- a(0) = 1 + 1 + 1 + ... = e − 1
- 1! 2! 3!
- a(1) = e − 1 + 1 + 1 + ... = 2e − 3
- 1! 2! 3!
- a(2) = 2e − 3 + e − 1 + 1 + ... = 7 e − 6
- 1! 2! 3! 2
- with e = 2.7182818... being Euler's constant.
- It can be shown that a(n) is of A(n) e + B(n) for integers A(n) and B(n).
- the form n!
- For example a(10) = 328161643 e − 652694486 .
- 10!
- Find A(10^9) + B(10^9) and give your answer mod 77 777 777.
- p_330_formula.gif
- Answer: 15955822
- Problem 331
- ===========
- N×N disks are placed on a square game board. Each disk has a black side
- and white side.
- At each turn, you may choose a disk and flip all the disks in the same row
- and the same column as this disk: thus 2×N-1 disks are flipped. The game
- ends when all disks show their white side. The following example shows a
- game on a 5×5 board.
- It can be proven that 3 is the minimal number of turns to finish this
- game.
- The bottom left disk on the N×N board has coordinates (0,0);
- the bottom right disk has coordinates (N-1,0) and the top left disk has
- coordinates (0,N-1).
- Let C[N] be the following configuration of a board with N×N disks:
- A disk at (x,y) satisfying , shows its black side; otherwise, it shows its
- white side. C[5] is shown above.
- Let T(N) be the minimal number of turns to finish a game starting from
- configuration C[N] or 0 if configuration C[N] is unsolvable.
- We have shown that T(5)=3. You are also given that T(10)=29 and T(1
- 000)=395253.
- Find .
- p_331_crossflips3.gif
- p_331_crossflips1.gif
- p_331_crossflips2.gif
- Answer: b609ccc578e71db9de0524fff94e1b70
- Problem 332
- ===========
- A spherical triangle is a figure formed on the surface of a sphere by
- three great circular arcs intersecting pairwise in three vertices.
- Let C(r) be the sphere with the centre (0,0,0) and radius r.
- Let Z(r) be the set of points on the surface of C(r) with integer
- coordinates.
- Let T(r) be the set of spherical triangles with vertices in
- Z(r).Degenerate spherical triangles, formed by three points on the same
- great arc, are not included in T(r).
- Let A(r) be the area of the smallest spherical triangle in T(r).
- For example A(14) is 3.294040 rounded to six decimal places.
- Find A(r). Give your answer rounded to six decimal places.
- p_332_spherical.jpg
- p_332_sum.gif
- Answer: c2ae53ebfb15db373cfe5d71078ea1ca
- Problem 333
- ===========
- All positive integers can be partitioned in such a way that each and every
- term of the partition can be expressed as 2^ix3^j, where i,j ≥ 0.
- Let's consider only those such partitions where none of the terms can
- divide any of the other terms.
- For example, the partition of 17 = 2 + 6 + 9 = (2^1x3^0 + 2^1x3^1 +
- 2^0x3^2) would not be valid since 2 can divide 6. Neither would the
- partition 17 = 16 + 1 = (2^4x3^0 + 2^0x3^0) since 1 can divide 16. The
- only valid partition of 17 would be 8 + 9 = (2^3x3^0 + 2^0x3^2).
- Many integers have more than one valid partition, the first being 11
- having the following two partitions.
- 11 = 2 + 9 = (2^1x3^0 + 2^0x3^2)
- 11 = 8 + 3 = (2^3x3^0 + 2^0x3^1)
- Let's define P(n) as the number of valid partitions of n. For example,
- P(11) = 2.
- Let's consider only the prime integers q which would have a single valid
- partition such as P(17).
- The sum of the primes q <100 such that P(q)=1 equals 233.
- Find the sum of the primes q <1000000 such that P(q)=1.
- Answer: 3053105
- Problem 334
- ===========
- In Plato's heaven, there exist an infinite number of bowls in a straight
- line.
- Each bowl either contains some or none of a finite number of beans.
- A child plays a game, which allows only one kind of move: removing two
- beans from any bowl, and putting one in each of the two adjacent bowls.
- The game ends when each bowl contains either one or no beans.
- For example, consider two adjacent bowls containing 2 and 3 beans
- respectively, all other bowls being empty. The following eight moves will
- finish the game:
- You are given the following sequences:
- t[0] = 123456.
- t[i-1] , if t[i-1] is even
- t[i] = 2
- t[i-1] 926252, if t[i-1] is odd
- 2
- where ⌊x⌋ is the floor function
- and is the bitwise XOR operator.
- b[i] = ( t[i] mod 2^11) + 1.
- The first two terms of the last sequence are b[1] = 289 and b[2] = 145.
- If we start with b[1] and b[2] beans in two adjacent bowls, 3419100 moves
- would be required to finish the game.
- Consider now 1500 adjacent bowls containing b[1], b[2],..., b[1500] beans
- respectively, all other bowls being empty. Find how many moves it takes
- before the game ends.
- p_334_beans.gif
- p_334_cases.gif
- p_334_lfloor.gif
- p_334_rfloor.gif
- p_334_oplus.gif
- Answer: 71851da3058acf6b74e90251bdf4aa8f
- Problem 335
- ===========
- Whenever Peter feels bored, he places some bowls, containing one bean
- each, in a circle. After this, he takes all the beans out of a certain
- bowl and drops them one by one in the bowls going clockwise. He repeats
- this, starting from the bowl he dropped the last bean in, until the
- initial situation appears again. For example with 5 bowls he acts as
- follows:
- So with 5 bowls it takes Peter 15 moves to return to the initial
- situation.
- Let M(x) represent the number of moves required to return to the initial
- situation, starting with x bowls. Thus, M(5) = 15. It can also be verified
- that M(100) = 10920.
- Find M(2^k+1). Give your answer modulo 7^9.
- p_335_mancala.gif
- p_335_sum.gif
- Answer: 5032316
- Problem 336
- ===========
- A train is used to transport four carriages in the order: ABCD. However,
- sometimes when the train arrives to collect the carriages they are not in
- the correct order.
- To rearrange the carriages they are all shunted on to a large rotating
- turntable. After the carriages are uncoupled at a specific point the train
- moves off the turntable pulling the carriages still attached with it. The
- remaining carriages are rotated 180 degrees. All of the carriages are then
- rejoined and this process is repeated as often as necessary in order to
- obtain the least number of uses of the turntable.
- Some arrangements, such as ADCB, can be solved easily: the carriages are
- separated between A and D, and after DCB are rotated the correct order has
- been achieved.
- However, Simple Simon, the train driver, is not known for his efficiency,
- so he always solves the problem by initially getting carriage A in the
- correct place, then carriage B, and so on.
- Using four carriages, the worst possible arrangements for Simon, which we
- shall call maximix arrangements, are DACB and DBAC; each requiring him
- five rotations (although, using the most efficient approach, they could be
- solved using just three rotations). The process he uses for DACB is shown
- below.
- It can be verified that there are 24 maximix arrangements for six
- carriages, of which the tenth lexicographic maximix arrangement is DFAECB.
- Find the 2011^th lexicographic maximix arrangement for eleven carriages.
- p_336_maximix.gif
- Answer: 7968e48fc692ce25bf7f5494f4ab6814
- Problem 337
- ===========
- Let {a[1], a[2],..., a[n]} be an integer sequence of length n such that:
- • a[1] = 6
- • for all 1 ≤ i < n : φ(a[i]) < φ(a[i+1]) < a[i] < a[i+1] ^1
- Let S(N) be the number of such sequences with a[n] ≤ N.
- For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}.
- We can verify that S(100) = 482073668 and S(10 000) mod 10^8 = 73808307.
- Find S(20 000 000) mod 10^8.
- ^1 φ denotes Euler's totient function.
- Answer: 85068035
- Problem 338
- ===========
- A rectangular sheet of grid paper with integer dimensions w × h is given.
- Its grid spacing is 1.
- When we cut the sheet along the grid lines into two pieces and rearrange
- those pieces without overlap, we can make new rectangles with different
- dimensions.
- For example, from a sheet with dimensions 9 × 4 , we can make rectangles
- with dimensions 18 × 2, 12 × 3 and 6 × 6 by cutting and rearranging as
- below:
- Similarly, from a sheet with dimensions 9 × 8 , we can make rectangles
- with dimensions 18 × 4 and 12 × 6 .
- For a pair w and h, let F(w,h) be the number of distinct rectangles that
- can be made from a sheet with dimensions w × h .
- For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2.
- Note that rectangles congruent to the initial one are not counted in
- F(w,h).
- Note also that rectangles with dimensions w × h and dimensions h × w are
- not considered distinct.
- For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h
- which satisfy 0 < h ≤ w ≤ N.
- We can verify that G(10) = 55, G(10^3) = 971745 and G(10^5) = 9992617687.
- Find G(10^12). Give your answer modulo 10^8.
- p_338_gridpaper.gif
- Answer: 15614292
- Problem 339
- ===========
- "And he came towards a valley, through which ran a river; and the borders
- of the valley were wooded, and on each side of the river were level
- meadows. And on one side of the river he saw a flock of white sheep, and
- on the other a flock of black sheep. And whenever one of the white sheep
- bleated, one of the black sheep would cross over and become white; and
- when one of the black sheep bleated, one of the white sheep would cross
- over and become black."
- [1]en.wikisource.org
- Initially each flock consists of n sheep. Each sheep (regardless of
- colour) is equally likely to be the next sheep to bleat. After a sheep has
- bleated and a sheep from the other flock has crossed over, Peredur may
- remove a number of white sheep in order to maximize the expected final
- number of black sheep. Let E(n) be the expected final number of black
- sheep if Peredur uses an optimal strategy.
- You are given that E(5) = 6.871346 rounded to 6 places behind the decimal
- point.
- Find E(10 000) and give your answer rounded to 6 places behind the decimal
- point.
- Visible links
- 1. http://en.wikisource.org/wiki/The_Mabinogion/Peredur_the_Son_of_Evrawc
- Answer: 0be02210b2d2212d37d026478093c457
- Problem 340
- ===========
- For fixed integers a, b, c, define the crazy function F(n) as follows:
- F(n) = n - c for all n > b
- F(n) = F(a + F(a + F(a + F(a + n)))) for all n ≤ b.
- Also, define S(a, b, c) = .
- For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000)
- = 2040.
- Also, S(50, 2000, 40) = 5204240.
- Find the last 9 digits of S(21^7, 7^21, 12^7).
- p_340_formula.gif
- Answer: fc838afe9ecde39bbe230923d7b50775
- Problem 341
- ===========
- The Golomb's self-describing sequence {G(n)} is the only nondecreasing
- sequence of natural numbers such that n appears exactly G(n) times in the
- sequence. The values of G(n) for the first few n are
- n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 …
- G(n) 1 2 2 3 3 4 4 4 5 5 5 6 6 6 6 …
- You are given that G(10^3) = 86, G(10^6) = 6137.
- You are also given that ΣG(n^3) = 153506976 for 1 ≤ n < 10^3.
- Find ΣG(n^3) for 1 ≤ n < 10^6.
- Answer: 7c163c3b4886943667b5c89db0a6cd02
- Problem 342
- ===========
- Consider the number 50.
- 50^2 = 2500 = 2^2 × 5^4, so φ(2500) = 2 × 4 × 5^3 = 8 × 5^3 = 2^3 × 5^3.
- ^1
- So 2500 is a square and φ(2500) is a cube.
- Find the sum of all numbers n, 1 < n < 10^10 such that φ(n^2) is a cube.
- ^1 φ denotes Euler's totient function.
- Answer: 0e9add0383d4116c7c5cb3dc73fc0536
- Problem 343
- ===========
- For any positive integer k, a finite sequence a[i] of fractions x[i]/y[i]
- is defined by:
- a[1] = 1/k and
- a[i] = (x[i-1]+1)/(y[i-1]-1) reduced to lowest terms for i>1.
- When a[i] reaches some integer n, the sequence stops. (That is, when
- y[i]=1.)
- Define f(k) = n.
- For example, for k = 20:
- 1/20 → 2/19 → 3/18 = 1/6 → 2/5 → 3/4 → 4/3 → 5/2 → 6/1 = 6
- So f(20) = 6.
- Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k^3) = 118937 for 1 ≤ k ≤ 100.
- Find Σf(k^3) for 1 ≤ k ≤ 2×10^6.
- Answer: 0e10bd111425ad8e1343ac79dac7bb0e
- Problem 344
- ===========
- One variant of N.G. de Bruijn's silver dollar game can be described as
- follows:
- On a strip of squares a number of coins are placed, at most one coin per
- square. Only one coin, called the silver dollar, has any value. Two
- players take turns making moves. At each turn a player must make either a
- regular or a special move.
- A regular move consists of selecting one coin and moving it one or more
- squares to the left. The coin cannot move out of the strip or jump on or
- over another coin.
- Alternatively, the player can choose to make the special move of pocketing
- the leftmost coin rather than making a regular move. If no regular moves
- are possible, the player is forced to pocket the leftmost coin.
- The winner is the player who pockets the silver dollar.
- A winning configuration is an arrangement of coins on the strip where the
- first player can force a win no matter what the second player does.
- Let W(n,c) be the number of winning configurations for a strip of n
- squares, c worthless coins and one silver dollar.
- You are given that W(10,2) = 324 and W(100,10) = 1514704946113500.
- Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003
- · 1 000 033).
- p_344_silverdollar.gif
- Answer: 38e7b980b38fcac89b3e267e328cd292
- Problem 345
- ===========
- We define the Matrix Sum of a matrix as the maximum sum of matrix elements
- with each element being the only one in his row and column. For example,
- the Matrix Sum of the matrix below equals 3315 ( = 863 + 383 + 343 + 959 +
- 767):
- 7 53 183 439 863
- 497 383 563 79 973
- 287 63 343 169 583
- 627 343 773 959 943
- 767 473 103 699 303
- Find the Matrix Sum of:
- 7 53 183 439 863 497 383 563 79 973 287 63 343 169 583
- 627 343 773 959 943 767 473 103 699 303 957 703 583 639 913
- 447 283 463 29 23 487 463 993 119 883 327 493 423 159 743
- 217 623 3 399 853 407 103 983 89 463 290 516 212 462 350
- 960 376 682 962 300 780 486 502 912 800 250 346 172 812 350
- 870 456 192 162 593 473 915 45 989 873 823 965 425 329 803
- 973 965 905 919 133 673 665 235 509 613 673 815 165 992 326
- 322 148 972 962 286 255 941 541 265 323 925 281 601 95 973
- 445 721 11 525 473 65 511 164 138 672 18 428 154 448 848
- 414 456 310 312 798 104 566 520 302 248 694 976 430 392 198
- 184 829 373 181 631 101 969 613 840 740 778 458 284 760 390
- 821 461 843 513 17 901 711 993 293 157 274 94 192 156 574
- 34 124 4 878 450 476 712 914 838 669 875 299 823 329 699
- 815 559 813 459 522 788 168 586 966 232 308 833 251 631 107
- 813 883 451 509 615 77 281 613 459 205 380 274 302 35 805
- Answer: 13938
- Problem 346
- ===========
- The number 7 is special, because 7 is 111 written in base 2, and 11
- written in base 6
- (i.e. 7[10] = 11[6] = 111[2]). In other words, 7 is a repunit in at least
- two bases b > 1.
- We shall call a positive integer with this property a strong repunit. It
- can be verified that there are 8 strong repunits below 50:
- {1,7,13,15,21,31,40,43}.
- Furthermore, the sum of all strong repunits below 1000 equals 15864.
- Find the sum of all strong repunits below 10^12.
- Answer: a17874b5a9ec9d7fc8c6489ab8ff29b9
- Problem 347
- ===========
- The largest integer ≤ 100 that is only divisible by both the primes 2 and
- 3 is 96, as 96=32*3=2^5*3.For two distinct primes p and q let M(p,q,N) be
- the largest positive integer ≤N only divisibleby both p and q and
- M(p,q,N)=0 if such a positive integer does not exist.
- E.g. M(2,3,100)=96.
- M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5.
- Also M(2,73,100)=0 because there does not exist a positive integer ≤ 100
- that is divisible by both 2 and 73.
- Let S(N) be the sum of all distinct M(p,q,N).S(100)=2262.
- Find S(10 000 000).
- Answer: 96ce0eabcbe7a2b2eb1197a1bcc5d37b
- Problem 348
- ===========
- Many numbers can be expressed as the sum of a square and a cube. Some of
- them in more than one way.
- Consider the palindromic numbers that can be expressed as the sum of a
- square and a cube, both greater than 1, in exactly 4 different ways.
- For example, 5229225 is a palindromic number and it can be expressed in
- exactly 4 different ways:
- 2285^2 + 20^3
- 2223^2 + 66^3
- 1810^2 + 125^3
- 1197^2 + 156^3
- Find the sum of the five smallest such palindromic numbers.
- Answer: f286f9159fc20aeb97a8bf8396ba64de
- Problem 349
- ===========
- An ant moves on a regular grid of squares that are coloured either black
- or white.
- The ant is always oriented in one of the cardinal directions (left, right,
- up or down) and moves from square to adjacent square according to the
- following rules:
- - if it is on a black square, it flips the color of the square to white,
- rotates 90 degrees counterclockwise and moves forward one square.
- - if it is on a white square, it flips the color of the square to black,
- rotates 90 degrees clockwise and moves forward one square.
- Starting with a grid that is entirely white, how many squares are black
- after 10^18 moves of the ant?
- Answer: 412b0faec10b3adb415363d2df26530d
- Problem 350
- ===========
- A list of size n is a sequence of n natural numbers.
- Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11).
- The greatest common divisor, or gcd, of a list is the largest natural
- number that divides all entries of the list.
- Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11.
- The least common multiple, or lcm, of a list is the smallest natural
- number divisible by each entry of the list.
- Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11.
- Let f(G, L, N) be the number of lists of size N with gcd ≥ G and lcm ≤ L.
- For example:
- f(10, 100, 1) = 91.
- f(10, 100, 2) = 327.
- f(10, 100, 3) = 1135.
- f(10, 100, 1000) mod 101^4 = 3286053.
- Find f(10^6, 10^12, 10^18) mod 101^4.
- Answer: 84664213
- Problem 351
- ===========
- A hexagonal orchard of order n is a triangular lattice made up of points
- within a regular hexagon with side n. The following is an example of a
- hexagonal orchard of order 5:
- Highlighted in green are the points which are hidden from the center by a
- point closer to it. It can be seen that for a hexagonal orchard of order
- 5, 30 points are hidden from the center.
- Let H(n) be the number of points hidden from the center in a hexagonal
- orchard of order n.
- H(5) = 30. H(10) = 138. H(1 000) = 1177848.
- Find H(100 000 000).
- p_351_hexorchard.png
- Answer: 338481092e945257756075a8d03978fd
- Problem 352
- ===========
- Each one of the 25 sheep in a flock must be tested for a rare virus, known
- to affect 2% of the sheep population.An accurate and extremely sensitive
- PCR test exists for blood samples, producing a clear positive / negative
- result, but it is very time-consuming and expensive.
- Because of the high cost, the vet-in-charge suggests that instead of
- performing 25 separate tests, the following procedure can be used instead:
- The sheep are split into 5 groups of 5 sheep in each group. For each
- group, the 5 samples are mixed together and a single test is performed.
- Then,
- • If the result is negative, all the sheep in that group are deemed to
- be virus-free.
- • If the result is positive, 5 additional tests will be performed (a
- separate test for each animal) to determine the affected
- individual(s).
- Since the probability of infection for any specific animal is only 0.02,
- the first test (on the pooled samples) for each group will be:
- • Negative (and no more tests needed) with probability 0.98^5 =
- 0.9039207968.
- • Positive (5 additional tests needed) with probability 1 - 0.9039207968
- = 0.0960792032.
- Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5
- = 1.480396016.
- Consequently, all 5 groups can be screened using an average of only
- 1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of more
- than 70% !
- Although the scheme we have just described seems to be very efficient, it
- can still be improved considerably (always assuming that the test is
- sufficiently sensitive and that there are no adverse effects caused by
- mixing different samples). E.g.:
- • We may start by running a test on a mixture of all the 25 samples. It
- can be verified that in about 60.35% of the cases this test will be
- negative, thus no more tests will be needed. Further testing will only
- be required for the remaining 39.65% of the cases.
- • If we know that at least one animal in a group of 5 is infected and
- the first 4 individual tests come out negative, there is no need to
- run a test on the fifth animal (we know that it must be infected).
- • We can try a different number of groups / different number of animals
- in each group, adjusting those numbers at each level so that the total
- expected number of tests will be minimised.
- To simplify the very wide range of possibilities, there is one restriction
- we place when devising the most cost-efficient testing scheme: whenever we
- start with a mixed sample, all the sheep contributing to that sample must
- be fully screened (i.e. a verdict of infected / virus-free must be reached
- for all of them) before we start examining any other animals.
- For the current example, it turns out that the most cost-efficient testing
- scheme (we'll call it the optimal strategy) requires an average of just
- 4.155452 tests!
- Using the optimal strategy, let T(s,p) represent the average number of
- tests needed to screen a flock of s sheep for a virus having probability p
- to be present in any individual.
- Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25,
- 0.10) = 12.702124.
- Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50.
- Give your answer rounded to six decimal places.
- Answer: 2e74b2fb574d6318cdbf2a41ad006de7
- Problem 353
- ===========
- A moon could be described by the sphere C(r) with centre (0,0,0) and
- radius r.
- There are stations on the moon at the points on the surface of C(r) with
- integer coordinates. The station at (0,0,r) is called North Pole station,
- the station at (0,0,-r) is called South Pole station.
- All stations are connected with each other via the shortest road on the
- great arc through the stations. A journey between two stations is risky.
- If d is the length of the road between two stations, (d/(π r))^2 is a
- measure for the risk of the journey (let us call it the risk of the road).
- If the journey includes more than two stations, the risk of the journey is
- the sum of risks of the used roads.
- A direct journey from the North Pole station to the South Pole station has
- the length πr and risk 1. The journey from the North Pole station to the
- South Pole station via (0,r,0) has the same length, but a smaller risk:
- (½πr/(πr))^2+(½πr/(πr))^2=0.5.
- The minimal risk of a journey from the North Pole station to the South
- Pole station on C(r) is M(r).
- You are given that M(7)=0.1784943998 rounded to 10 digits behind the
- decimal point.
- Find ∑M(2^n-1) for 1≤n≤15.
- Give your answer rounded to 10 digits behind the decimal point in the form
- a.bcdefghijk.
- Answer: 211b5626459be71baefc78478d18bdc3
- Problem 354
- ===========
- Consider a honey bee's honeycomb where each cell is a perfect regular
- hexagon with side length 1.
- One particular cell is occupied by the queen bee.
- For a positive real number L, let B(L) count the cells with distance L
- from the queen bee cell (all distances are measured from centre to
- centre); you may assume that the honeycomb is large enough to accommodate
- for any distance we wish to consider.
- For example, B(√3) = 6, B(√21) = 12 and B(111 111 111) = 54.
- Find the number of L ≤ 5·10^11 such that B(L) = 450.
- p_354_bee_honeycomb.png
- Answer: 58065134
- Problem 355
- ===========
- Define Co(n) to be the maximal possible sum of a set of mutually co-prime
- elements from {1, 2, ..., n}.
- For example Co(10) is 30 and hits that maximum on the subset
- {1, 5, 7, 8, 9}.
- You are given that Co(30) = 193 and Co(100) = 1356.
- Find Co(200000).
- Answer: 41cb97b6d02878d79f8b2e3b6c74920a
- Problem 356
- ===========
- Let a[n] be the largest real root of a polynomial g(x) = x^3 - 2^n·x^2 +
- n.
- For example, a[2] = 3.86619826...
- Find the last eight digits of.
- Note: represents the floor function.
- p_356_cubicpoly1.gif
- p_356_cubicpoly2.gif
- Answer: 28010159
- Problem 357
- ===========
- Consider the divisors of 30: 1,2,3,5,6,10,15,30.
- It can be seen that for every divisor d of 30, d+30/d is prime.
- Find the sum of all positive integers n not exceeding 100 000 000
- such thatfor every divisor d of n, d+n/d is prime.
- Answer: ed25b13b18a21c1077fed00ef42f503b
- Problem 358
- ===========
- A cyclic number with n digits has a very interesting property:
- When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly
- the same digits, in the same order, but rotated in a circular fashion!
- The smallest cyclic number is the 6-digit number 142857 :
- 142857 × 1 = 142857
- 142857 × 2 = 285714
- 142857 × 3 = 428571
- 142857 × 4 = 571428
- 142857 × 5 = 714285
- 142857 × 6 = 857142
- The next cyclic number is 0588235294117647 with 16 digits :
- 0588235294117647 × 1 = 0588235294117647
- 0588235294117647 × 2 = 1176470588235294
- 0588235294117647 × 3 = 1764705882352941
- ...
- 0588235294117647 × 16 = 9411764705882352
- Note that for cyclic numbers, leading zeros are important.
- There is only one cyclic number for which, the eleven leftmost digits are
- 00000000137 and the five rightmost digits are 56789 (i.e., it has the form
- 00000000137...56789 with an unknown number of digits in the middle). Find
- the sum of all its digits.
- Answer: 359e1ec8aeaa3932b54f2a5d20fa4f73
- Problem 359
- ===========
- An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get
- a room at Hilbert's newest infinite hotel. The hotel contains an infinite
- number of floors (numbered 1, 2, 3, etc.), and each floor contains an
- infinite number of rooms (numbered 1, 2, 3, etc.).
- Initially the hotel is empty. Hilbert declares a rule on how the n^th
- person is assigned a room: person n gets the first vacant room in the
- lowest numbered floor satisfying either of the following:
- • the floor is empty
- • the floor is not empty, and if the latest person taking a room in that
- floor is person m, then m + n is a perfect square
- Person 1 gets room 1 in floor 1 since floor 1 is empty.
- Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect
- square.
- Person 2 instead gets room 1 in floor 2 since floor 2 is empty.
- Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square.
- Eventually, every person in the line gets a room in the hotel.
- Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no
- person occupies the room. Here are a few examples:
- P(1, 1) = 1
- P(1, 2) = 3
- P(2, 1) = 2
- P(10, 20) = 440
- P(25, 75) = 4863
- P(99, 100) = 19454
- Find the sum of all P(f, r) for all positive f and r such that f × r =
- 71328803586048 and give the last 8 digits as your answer.
- Answer: 40632119
- Problem 360
- ===========
- Given two points (x[1],y[1],z[1]) and (x[2],y[2],z[2]) in three
- dimensional space, the Manhattan distance between those points is defined
- as
- |x[1]-x[2]|+|y[1]-y[2]|+|z[1]-z[2]|.
- Let C(r) be a sphere with radius r and center in the origin O(0,0,0).
- Let I(r) be the set of all points with integer coordinates on the surface
- of C(r).
- Let S(r) be the sum of the Manhattan distances of all elements of I(r) to
- the origin O.
- E.g. S(45)=34518.
- Find S(10^10).
- Answer: 82ec91527315eafb7e3acc139eeeb8eb
- Problem 361
- ===========
- The Thue-Morse sequence {T[n]} is a binary sequence satisfying:
- • T[0] = 0
- • T[2n] = T[n]
- • T[2n+1] = 1 - T[n]
- The first several terms of {T[n]} are given as follows:
- 01101001100101101001011001101001....
- We define {A[n]} as the sorted sequence of integers such that the binary
- expression of each element appears as a subsequence in {T[n]}.
- For example, the decimal number 18 is expressed as 10010 in binary. 10010
- appears in {T[n]} (T[8] to T[12]), so 18 is an element of {A[n]}.
- The decimal number 14 is expressed as 1110 in binary. 1110 never appears
- in {T[n]}, so 14 is not an element of {A[n]}.
- The first several terms of A[n] are given as follows:
- n 0 1 2 3 4 5 6 7 8 9 10 11 12 …
- A[n] 0 1 2 3 4 5 6 9 10 11 12 13 18 …
- We can also verify that A[100] = 3251 and A[1000] = 80852364498.
- Find the last 9 digits of .
- p_361_Thue-Morse1.gif
- Answer: 178476944
- Problem 362
- ===========
- Consider the number 54.
- 54 can be factored in 7 distinct ways into one or more factors larger than
- 1:
- 54, 2×27, 3×18, 6×9, 3×3×6, 2×3×9 and 2×3×3×3.
- If we require that the factors are all squarefree only two ways remain:
- 3×3×6 and 2×3×3×3.
- Let's call Fsf(n) the number of ways n can be factored into one or more
- squarefree factors larger than 1, soFsf(54)=2.
- Let S(n) be ∑Fsf(k) for k=2 to n.
- S(100)=193.
- Find S(10 000 000 000).
- Answer: b62f0d524bec8653ba7b8a2cab70260b
- Problem 363
- ===========
- A cubic Bézier curve is defined by four points: P[0], P[1], P[2] and P[3].
- The curve is constructed as follows:
- On the segments P[0]P[1], P[1]P[2] and P[2]P[3] the points Q[0],Q[1] and
- Q[2] are drawn such that
- P[0]Q[0]/P[0]P[1]=P[1]Q[1]/P[1]P[2]=P[2]Q[2]/P[2]P[3]=t (t in [0,1]).
- On the segments Q[0]Q[1] and Q[1]Q[2] the points R[0] and R[1] are drawn
- such thatQ[0]R[0]/Q[0]Q[1]=Q[1]R[1]/Q[1]Q[2]=t for the same value of t.
- On the segment R[0]R[1] the point B is drawn such that R[0]B/R[0]R[1]=t
- for the same value of t.The Bézier curve defined by the points P[0], P[1],
- P[2], P[3] is the locus of B as Q[0] takes all possible positions on the
- segment P[0]P[1]. (Please note that for all points the value of t is the
- same.)
- [1]Applet
- In the applet to the right you can drag the points P[0], P[1], P[2] and
- P[3] to see what the Bézier curve (green curve) defined by those points
- looks like. You can also drag the point Q[0] along the segment P[0]P[1].
- From the construction it is clear that the Bézier curve will be tangent to
- the segments P[0]P[1] in P[0] and P[2]P[3] in P[3].
- A cubic Bézier curve with P[0]=(1,0), P[1]=(1,v), P[2]=(v,1) and
- P[3]=(0,1) is used to approximate a quarter circle.
- The value v>0 is chosen such that the area enclosed by the lines OP[0],
- OP[3] and the curve is equal to ^π/[4] (the area of the quarter circle).
- By how many percent does the length of the curve differ from the length of
- the quarter circle?
- That is, if L is the length of the curve, calculate 100*^(L-π/2)/[(π/2)].
- Give your answer rounded to 10 digits behind the decimal point.
- Visible links
- 1. CabriJava.class
- Answer: 2bc63386b7cccc64c67f90e719936143
- Problem 364
- ===========
- There are N seats in a row. N people come after each other to fill the
- seats according to the following rules:
- 1. If there is any seat whose adjacent seat(s) are not occupied take such
- a seat.
- 2. If there is no such seat and there is any seat for which only one
- adjacent seat is occupied take such a seat.
- 3. Otherwise take one of the remaining available seats.
- Let T(N) be the number of possibilities that N seats are occupied by N
- people with the given rules.
- The following figure shows T(4)=8.
- We can verify that T(10) = 61632 and T(1 000) mod 100 000 007 = 47255094.
- Find T(1 000 000) mod 100 000 007.
- p_364_comf_dist.gif
- Answer: 44855254
- Problem 365
- ===========
- The binomial coeffient C(10^18,10^9) is a number with more than 9 billion
- (9×10^9) digits.
- Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m.
- Calculate ∑M(10^18,10^9,p*q*r) for 1000<p<q<r<5000 and p,q,r prime.
- Answer: 53addf69042b0cefbeb94f3bd3224918
- Problem 366
- ===========
- Two players, Anton and Bernhard, are playing the following game.
- There is one pile of n stones.
- The first player may remove any positive number of stones, but not the
- whole pile.
- Thereafter, each player may remove at most twice the number of stones his
- opponent took on the previous move.
- The player who removes the last stone wins.
- E.g. n=5
- If the first player takes anything more than one stone the next player
- will be able to take all remaining stones.
- If the first player takes one stone, leaving four, his opponent will take
- also one stone, leaving three stones.
- The first player cannot take all three because he may take at most 2x1=2
- stones. So let's say he takes also one stone, leaving 2. The second player
- can take the two remaining stones and wins.
- So 5 is a losing position for the first player.
- For some winning positions there is more than one possible move for the
- first player.
- E.g. when n=17 the first player can remove one or four stones.
- Let M(n) be the maximum number of stones the first player can take from a
- winning position at his first turn and M(n)=0 for any other position.
- ∑M(n) for n≤100 is 728.
- Find ∑M(n) for n≤10^18.Give your answer modulo 10^8.
- Answer: 88351299
- Problem 367
- ===========
- Bozo sort, not to be confused with the slightly less efficient bogo sort,
- consists out of checking if the input sequence is sorted and if not
- swapping randomly two elements. This is repeated until eventually the
- sequence is sorted.
- If we consider all permutations of the first 4 natural numbers as input
- the expectation value of the number of swaps, averaged over all 4! input
- sequences is 24.75.
- The already sorted sequence takes 0 steps.
- In this problem we consider the following variant on bozo sort.
- If the sequence is not in order we pick three elements at random and
- shuffle these three elements randomly.
- All 3!=6 permutations of those three elements are equally likely.
- The already sorted sequence will take 0 steps.
- If we consider all permutations of the first 4 natural numbers as input
- the expectation value of the number of shuffles, averaged over all 4!
- input sequences is 27.5.
- Consider as input sequences the permutations of the first 11 natural
- numbers.
- Averaged over all 11! input sequences, what is the expected number of
- shuffles this sorting algorithm will perform?
- Give your answer rounded to the nearest integer.
- Answer: 48271207
- Problem 368
- ===========
- The harmonic series 1 + 1 + 1 + 1 + ... is well known to be divergent.
- 2 3 4
- If we however omit from this series every term where the denominator has a
- 9 in it, the series remarkably enough converges to approximately
- 22.9206766193.
- This modified harmonic series is called the Kempner series.
- Let us now consider another modified harmonic series by omitting from the
- harmonic series every term where the denominator has 3 or more equal
- consecutive digits.One can verify that out of the first 1200 terms of the
- harmonic series, only 20 terms will be omitted.
- These 20 omitted terms are:
- 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 ,
- 111 222 333 444 555 666 777 888 999 1000 1110
- 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 and 1 .
- 1111 1112 1113 1114 1115 1116 1117 1118 1119
- This series converges as well.
- Find the value the series converges to.
- Give your answer rounded to 10 digits behind the decimal point.
- Answer: bfb15c388f4721cbd5eb89f17be2eef2
- Problem 369
- ===========
- In a standard 52 card deck of playing cards, a set of 4 cards is a Badugi
- if it contains 4 cards with no pairs and no two cards of the same suit.
- Let f(n) be the number of ways to choose n cards with a 4 card subset that
- is a Badugi. For example, there are 2598960 ways to choose five cards from
- a standard 52 card deck, of which 514800 contain a 4 card subset that is a
- Badugi, so f(5) = 514800.
- Find ∑f(n) for 4 ≤ n ≤ 13.
- Answer: 0f8828f58dbac4f15f296c79b686ed0e
- Problem 370
- ===========
- Let us define a geometric triangle as an integer sided triangle with sides
- a ≤ b ≤ c so that its sides form a geometric progression, i.e.
- b^2 = a · c .
- An example of such a geometric triangle is the triangle with sides a =
- 144, b = 156 and c = 169.
- There are 861805 geometric triangles with perimeter ≤ 10^6 .
- How many geometric triangles exist with perimeter ≤ 2.5·10^13 ?
- Answer: 85b5048e25677205555a5308991c2e04
- Problem 371
- ===========
- Oregon licence plates consist of three letters followed by a three digit
- number (each digit can be from [0..9]).
- While driving to work Seth plays the following game:
- Whenever the numbers of two licence plates seen on his trip add to 1000
- that's a win.
- E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as
- he sees them in the same trip).
- Find the expected number of plates he needs to see for a win.
- Give your answer rounded to 8 decimal places behind the decimal point.
- Note: We assume that each licence plate seen is equally likely to have any
- three digit number on it.
- Answer: 537403a97924621c604ce5ab6288b97d
- Problem 372
- ===========
- Let R(M, N) be the number of lattice points (x, y) which satisfy M<x≤N,
- M<y≤N and is odd.
- We can verify that R(0, 100) = 3019 and R(100, 10000) = 29750422.
- Find R(2·10^6, 10^9).
- Note: represents the floor function.
- p_372_pencilray1.jpg
- p_372_pencilray2.gif
- Answer: 5fdeda0dca23d12ae3eb1763b2c6f5ea
- Problem 373
- ===========
- Every triangle has a circumscribed circle that goes through the three
- vertices.Consider all integer sided triangles for which the radius of the
- circumscribed circle is integral as well.
- Let S(n) be the sum of the radii of the circumscribed circles of all such
- triangles for which the radius does not exceed n.
- S(100)=4950 and S(1200)=1653605.
- Find S(10^7).
- Answer: 888d60a6b2b4b9146d7c9c14ffd82673
- Problem 374
- ===========
- An integer partition of a number n is a way of writing n as a sum of
- positive integers.
- Partitions that differ only in the order of their summands are considered
- the same.A partition of n into distinct parts is a partition of n in which
- every part occurs at most once.
- The partitions of 5 into distinct parts are:
- 5, 4+1 and 3+2.
- Let f(n) be the maximum product of the parts of any such partition of n
- into distinct parts and let m(n) be the number of elements of any such
- partition of n with that product.
- So f(5)=6 and m(5)=2.
- For n=10 the partition with the largest product is 10=2+3+5, which gives
- f(10)=30 and m(10)=3.
- And their product, f(10)·m(10) = 30·3 = 90
- It can be verified that
- ∑f(n)·m(n) for 1 ≤ n ≤ 100 = 1683550844462.
- Find ∑f(n)·m(n) for 1 ≤ n ≤ 10^14.
- Give your answer modulo 982451653, the 50 millionth prime.
- Answer: 6fcb063062076b5aaaff3e3cd03e4b2f
- Problem 375
- ===========
- Let S[n] be an integer sequence produced with the following pseudo-random
- number generator:
- S[0] =[ ] 290797[ ]
- S[n+1] =[ ] S[n]^2 mod 50515093
- Let A(i, j) be the minimum of the numbers S[i], S[i+1], ... , S[j] for i ≤
- j.
- Let M(N) = ΣA(i, j) for 1 ≤ i ≤ j ≤ N.
- We can verify that M(10) = 432256955 and M(10 000) = 3264567774119.
- Find M(2 000 000 000).
- Answer: 68a12e3f2e4ccbae9c8555e547fbe096
- Problem 376
- ===========
- Consider the following set of dice with nonstandard pips:
- Die A: 1 4 4 4 4 4
- Die B: 2 2 2 5 5 5
- Die C: 3 3 3 3 3 6
- A game is played by two players picking a die in turn and rolling it. The
- player who rolls the highest value wins.
- If the first player picks die A and the second player picks die B we get
- P(second player wins) = 7/12 > 1/2
- If the first player picks die B and the second player picks die C we get
- P(second player wins) = 7/12 > 1/2
- If the first player picks die C and the second player picks die A we get
- P(second player wins) = 25/36 > 1/2
- So whatever die the first player picks, the second player can pick another
- die and have a larger than 50% chance of winning.
- A set of dice having this property is called a nontransitive set of dice.
- We wish to investigate how many sets of nontransitive dice exist. We will
- assume the following conditions:
- • There are three six-sided dice with each side having between 1 and N
- pips, inclusive.
- • Dice with the same set of pips are equal, regardless of which side on
- the die the pips are located.
- • The same pip value may appear on multiple dice; if both players roll
- the same value neither player wins.
- • The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set.
- For N = 7 we find there are 9780 such sets.
- How many are there for N = 30 ?
- Answer: c64df302990eb3738f8ec62ea6b66c0b
- Problem 377
- ===========
- There are 16 positive integers that do not have a zero in their digits and
- that have a digital sum equal to 5, namely:
- 5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111
- and 11111.
- Their sum is 17891.
- Let f(n) be the sum of all positive integers that do not have a zero in
- their digits and have a digital sum equal to n.
- Find .
- Give the last 9 digits as your answer.
- Answer: a915ccbae49de15208c88affba84d206
- Problem 378
- ===========
- Let T(n) be the n^th triangle number, so T(n) = n (n+1) .
- 2
- Let dT(n) be the number of divisors of T(n).
- E.g.:T(7) = 28 and dT(7) = 6.
- Let Tr(n) be the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n
- and dT(i) > dT(j) > dT(k).
- Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776.
- Find Tr(60 000 000).
- Give the last 18 digits of your answer.
- Answer: 336745dc9d90928596237c4b471a8927
- Problem 379
- ===========
- Let f(n) be the number of couples (x,y) with x and y positive integers, x
- ≤ y and the least common multiple of x and y equal to n.
- Let g be the summatory function of f, i.e.: g(n) = ∑ f(i) for 1 ≤ i ≤ n.
- You are given that g(10^6) = 37429395.
- Find g(10^12).
- Answer: de20f710cb6665c48795072197ad53e0
- Problem 380
- ===========
- An m×n maze is an m×n rectangular grid with walls placed between grid
- cells such that there is exactly one path from the top-left square to any
- other square.
- The following are examples of a 9×12 maze and a 15×20 maze:
- Let C(m,n) be the number of distinct m×n mazes. Mazes which can be formed
- by rotation and reflection from another maze are considered distinct.
- It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12)
- = 2.5720e46 (in scientific notation rounded to 5 significant digits).
- Find C(100,500) and write your answer in scientific notation rounded to 5
- significant digits.
- When giving your answer, use a lowercase e to separate mantissa and
- exponent.E.g. if the answer is 1234567891011 then the answer format would
- be 1.2346e12.
- Answer: c86d2f4c17c8134fbebed5d37a0f90d7
- Problem 381
- ===========
- For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5.
- For example, if p=7,
- (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! =
- 720+120+24+6+2 = 872.
- As 872 mod(7) = 4, S(7) = 4.
- It can be verified that ∑S(p) = 480 for 5 ≤ p < 100.
- Find ∑S(p) for 5 ≤ p < 10^8.
- Answer: 80c84973a9643e46d49d79d7284e7ff3
- Problem 382
- ===========
- A polygon is a flat shape consisting of straight line segments that are
- joined to form a closed chain or circuit. A polygon consists of at least
- three sides and does not self-intersect.
- A set S of positive numbers is said to generate a polygon P if:
- • no two sides of P are the same length,
- • the length of every side of P is in S, and
- • S contains no other value.
- For example:
- The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle).
- The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a
- quadrilateral).
- The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all.
- Consider the sequence s, defined as follows:
- • s[1] = 1, s[2] = 2, s[3] = 3
- • s[n] = s[n-1] + s[n-3] for n > 3.
- Let U[n] be the set {s[1], s[2], ..., s[n]}. For example, U[10] = {1, 2,
- 3, 4, 6, 9, 13, 19, 28, 41}.
- Let f(n) be the number of subsets of U[n] which generate at least one
- polygon.
- For example, f(5) = 7, f(10) = 501 and f(25) = 18635853.
- Find the last 9 digits of f(10^18).
- Answer: 56a121bcf3bb674d0d3ce561b6b24ea5
- Problem 383
- ===========
- Let f[5](n) be the largest integer x for which 5^x divides n.
- For example, f[5](625000) = 7.
- Let T[5](n) be the number of integers i which satisfy f[5]((2·i-1)!) <
- 2·f[5](i!) and 1 ≤ i ≤ n.
- It can be verified that T[5](10^3) = 68 and T[5](10^9) = 2408210.
- Find T[5](10^18).
- Answer: c1bc7c945344e1967bfaced9ade895a0
- Problem 384
- ===========
- Define the sequence a(n) as the number of adjacent pairs of ones in the
- binary expansion of n (possibly overlapping).
- E.g.: a(5) = a(101[2]) = 0, a(6) = a(110[2]) = 1, a(7) = a(111[2]) = 2
- Define the sequence b(n) = (-1)^a(n).
- This sequence is called the Rudin-Shapiro sequence.
- Also consider the summatory sequence of b(n): .
- The first couple of values of these sequences are:
- n 0 1 2 3 4 5 6 7
- a(n) 0 0 0 1 0 0 1 2
- b(n) 1 1 1 -1 1 1 -1 1
- s(n) 1 2 3 2 3 4 3 4
- The sequence s(n) has the remarkable property that all elements are
- positive and every positive integer k occurs exactly k times.
- Define g(t,c), with 1 ≤ c ≤ t, as the index in s(n) for which t occurs for
- the c'th time in s(n).
- E.g.: g(3,3) = 6, g(4,2) = 7 and g(54321,12345) = 1220847710.
- Let F(n) be the fibonacci sequence defined by:
- F(0)=F(1)=1 and
- F(n)=F(n-1)+F(n-2) for n>1.
- Define GF(t)=g(F(t),F(t-1)).
- Find ΣGF(t) for 2≤t≤45.
- Answer: ea0bb1fff1a51b48971762b93aeed103
- Problem 385
- ===========
- For any triangle T in the plane, it can be shown that there is a unique
- ellipse with largest area that is completely inside T.
- For a given n, consider triangles T such that:
- - the vertices of T have integer coordinates with absolute value ≤ n, and
- - the foci^1 of the largest-area ellipse inside T are (√13,0) and
- (-√13,0).
- Let A(n) be the sum of the areas of all such triangles.
- For example, if n = 8, there are two such triangles. Their vertices are
- (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each
- triangle is 36. Thus A(8) = 36 + 36 = 72.
- It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008.
- Find A(1 000 000 000).
- ^1The foci (plural of focus) of an ellipse are two points A and B such
- that for every point P on the boundary of the ellipse, AP + PB is
- constant.
- Answer: a21c033d9e119c293e51966ea78c9950
- Problem 386
- ===========
- Let n be an integer and S(n) be the set of factors of n.
- A subset A of S(n) is called an antichain of S(n) if A contains only one
- element or if none of the elements of A divides any of the other elements
- of A.
- For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30}
- {2, 5, 6} is not an antichain of S(30).
- {2, 3, 5} is an antichain of S(30).
- Let N(n) be the maximum length of an antichain of S(n).
- Find ΣN(n) for 1 ≤ n ≤ 10^8
- Answer: 528755790
- Problem 387
- ===========
- A Harshad or Niven number is a number that is divisible by the sum of its
- digits.
- 201 is a Harshad number because it is divisible by 3 (the sum of its
- digits.)
- When we truncate the last digit from 201, we get 20, which is a Harshad
- number.
- When we truncate the last digit from 20, we get 2, which is also a Harshad
- number.
- Let's call a Harshad number that, while recursively truncating the last
- digit, always results in a Harshad number a right truncatable Harshad
- number.
- Also:
- 201/3=67 which is prime.
- Let's call a Harshad number that, when divided by the sum of its digits,
- results in a prime a strong Harshad number.
- Now take the number 2011 which is prime.
- When we truncate the last digit from it we get 201, a strong Harshad
- number that is also right truncatable.
- Let's call such primes strong, right truncatable Harshad primes.
- You are given that the sum of the strong, right truncatable Harshad primes
- less than 10000 is 90619.
- Find the sum of the strong, right truncatable Harshad primes less than
- 10^14.
- Answer: a20cbd8639767decfa2c2c9955eb6be3
- Problem 388
- ===========
- Consider all lattice points (a,b,c) with 0 ≤ a,b,c ≤ N.
- From the origin O(0,0,0) all lines are drawn to the other lattice points.
- Let D(N) be the number of distinct such lines.
- You are given that D(1 000 000) = 831909254469114121.
- Find D(10^10). Give as your answer the first nine digits followed by the
- last nine digits.
- Answer: 2bab886c7d98d802d9249c9e12d72c25
- Problem 389
- ===========
- An unbiased single 4-sided die is thrown and its value, T, is noted.
- T unbiased 6-sided dice are thrown and their scores are added together.
- The sum, C, is noted.
- C unbiased 8-sided dice are thrown and their scores are added together.
- The sum, O, is noted.
- O unbiased 12-sided dice are thrown and their scores are added together.
- The sum, D, is noted.
- D unbiased 20-sided dice are thrown and their scores are added together.
- The sum, I, is noted.
- Find the variance of I, and give your answer rounded to 4 decimal places.
- Answer: 79a080d38b837547b975c97b44764dfb
- Problem 390
- ===========
- Consider the triangle with sides √5, √65 and √68.It can be shown that this
- triangle has area 9.
- S(n) is the sum of the areas of all triangles with sides √(1+b^2),
- √(1+c^2) and √(b^2+c^2) (for positive integers b and c ) that have an
- integral area not exceeding n.
- The example triangle has b=2 and c=8.
- S(10^6)=18018206.
- Find S(10^10).
- Answer: ed7f2fbc05a2fd2033d80de671f35ea3
- Problem 391
- ===========
- Let s[k] be the number of 1’s when writing the numbers from 0 to k in
- binary.
- For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101.
- There are seven 1’s, so s[5] = 7.
- The sequence S = {s[k] : k ≥ 0} starts {0, 1, 2, 4, 5, 7, 9, 12, ...}.
- A game is played by two players. Before the game starts, a number n is
- chosen. A counter c starts at 0. At each turn, the player chooses a number
- from 1 to n (inclusive) and increases c by that number. The resulting
- value of c must be a member of S. If there are no more valid moves, the
- player loses.
- For example:
- Let n = 5. c starts at 0.
- Player 1 chooses 4, so c becomes 0 + 4 = 4.
- Player 2 chooses 5, so c becomes 4 + 5 = 9.
- Player 1 chooses 3, so c becomes 9 + 3 = 12.
- etc.
- Note that c must always belong to S, and each player can increase c by at
- most n.
- Let M(n) be the highest number the first player can choose at her first
- turn to force a win, and M(n) = 0 if there is no such move. For example,
- M(2) = 2, M(7) = 1 and M(20) = 4.
- Given Σ(M(n))^3 = 8150 for 1 ≤ n ≤ 20.
- Find Σ(M(n))^3 for 1 ≤ n ≤ 1000.
- Answer: b2947548d4f5c4878c5f788f9849e750
- Problem 392
- ===========
- A rectilinear grid is an orthogonal grid where the spacing between the
- gridlines does not have to be equidistant.
- An example of such grid is logarithmic graph paper.
- Consider rectilinear grids in the Cartesian coordinate system with the
- following properties:
- • The gridlines are parallel to the axes of the Cartesian coordinate
- system.
- • There are N+2 vertical and N+2 horizontal gridlines. Hence there are
- (N+1) x (N+1) rectangular cells.
- • The equations of the two outer vertical gridlines are x = -1 and x =
- 1.
- • The equations of the two outer horizontal gridlines are y = -1 and y =
- 1.
- • The grid cells are colored red if they overlap with the unit circle,
- black otherwise.
- For this problem we would like you to find the postions of the remaining N
- inner horizontal and N inner vertical gridlines so that the area occupied
- by the red cells is minimized.
- E.g. here is a picture of the solution for N = 10:
- The area occupied by the red cells for N = 10 rounded to 10 digits behind
- the decimal point is 3.3469640797.
- Find the positions for N = 400.
- Give as your answer the area occupied by the red cells rounded to 10
- digits behind the decimal point.
- Answer: 3268b0bc489187db3d234c097040d909
- Problem 393
- ===========
- An n×n grid of squares contains n^2 ants, one ant per square.
- All ants decide to move simultaneously to an adjacent square (usually 4
- possibilities, except for ants on the edge of the grid or at the corners).
- We define f(n) to be the number of ways this can happen without any ants
- ending on the same square and without any two ants crossing the same edge
- between two squares.
- You are given that f(4) = 88.
- Find f(10).
- Answer: 58e4990838fb3d1725872da30f9db748
- Problem 394
- ===========
- Jeff eats a pie in an unusual way.
- The pie is circular. He starts with slicing an initial cut in the pie
- along a radius.
- While there is at least a given fraction F of pie left, he performs the
- following procedure:
- - He makes two slices from the pie centre to any point of what is
- remaining of the pie border, any point on the remaining pie border equally
- likely. This will divide the remaining pie into three pieces.
- - Going counterclockwise from the initial cut, he takes the first two pie
- pieces and eats them.
- When less than a fraction F of pie remains, he does not repeat this
- procedure. Instead, he eats all of the remaining pie.
- For x ≥ 1, let E(x) be the expected number of times Jeff repeats the
- procedure above with F = ^1/[x].
- It can be verified that E(1) = 1, E(2) ≈ 1.2676536759, and E(7.5) ≈
- 2.1215732071.
- Find E(40) rounded to 10 decimal places behind the decimal point.
- Answer: f8ad575e1a03365a60b6429c3b7a64df
- Problem 395
- ===========
- The Pythagorean tree is a fractal generated by the following procedure:
- Start with a unit square. Then, calling one of the sides its base (in the
- animation, the bottom side is the base):
- 1. Attach a right triangle to the side opposite the base, with the
- hypotenuse coinciding with that side and with the sides in a 3-4-5
- ratio. Note that the smaller side of the triangle must be on the
- 'right' side with respect to the base (see animation).
- 2. Attach a square to each leg of the right triangle, with one of its
- sides coinciding with that leg.
- 3. Repeat this procedure for both squares, considering as their bases the
- sides touching the triangle.
- The resulting figure, after an infinite number of iterations, is the
- Pythagorean tree.
- It can be shown that there exists at least one rectangle, whose sides are
- parallel to the largest square of the Pythagorean tree, which encloses the
- Pythagorean tree completely.
- Find the smallest area possible for such a bounding rectangle, and give
- your answer rounded to 10 decimal places.
- p_395_pythagorean.gif
- Answer: 505048b0c619161d05b9b3e492f3edc3
- Problem 396
- ===========
- For any positive integer n, the nth weak Goodstein sequence {g[1], g[2],
- g[3], ...} is defined as:
- • g[1] = n
- • for k > 1, g[k] is obtained by writing g[k-1] in base k, interpreting
- it as a base k + 1 number, and subtracting 1.
- The sequence terminates when g[k] becomes 0.
- For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}:
- • g[1] = 6.
- • g[2] = 11 since 6 = 110[2], 110[3] = 12, and 12 - 1 = 11.
- • g[3] = 17 since 11 = 102[3], 102[4] = 18, and 18 - 1 = 17.
- • g[4] = 25 since 17 = 101[4], 101[5] = 26, and 26 - 1 = 25.
- and so on.
- It can be shown that every weak Goodstein sequence terminates.
- Let G(n) be the number of nonzero elements in the nth weak Goodstein
- sequence.
- It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381.
- It can also be verified that ΣG(n) = 2517 for 1 ≤ n < 8.
- Find the last 9 digits of ΣG(n) for 1 ≤ n < 16.
- Answer: 173214653
- Problem 397
- ===========
- On the parabola y = x^2/k, three points A(a, a^2/k), B(b, b^2/k) and C(c,
- c^2/k) are chosen.
- Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such
- that at least one angle of the triangle ABC is 45-degree, with 1 ≤ k ≤ K
- and -X ≤ a < b < c ≤ X.
- For example, F(1, 10) = 41 and F(10, 100) = 12492.
- Find F(10^6, 10^9).
- Answer: 07f769df9543bc05e6318878c34d074d
- Problem 398
- ===========
- Inside a rope of length n, n-1 points are placed with distance 1 from each
- other and from the endpoints. Among these points, we choose m-1 points at
- random and cut the rope at these points to create m segments.
- Let E(n, m) be the expected length of the second-shortest segment.For
- example, E(3, 2) = 2 and E(8, 3) = 16/7.Note that if multiple segments
- have the same shortest length the length of the second-shortest segment is
- defined as the same as the shortest length.
- Find E(10^7, 100).Give your answer rounded to 5 decimal places behind the
- decimal point.
- Answer: fa0a25d62fa225e05fd8736713a9bfc0
- Problem 399
- ===========
- The first 15 fibonacci numbers are:
- 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610.
- It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and
- 144 is divisible by 4 and by 9.
- So the first 13 squarefree fibonacci numbers are:
- 1,1,2,3,5,13,21,34,55,89,233,377 and 610.
- The 200th squarefree fibonacci number
- is:971183874599339129547649988289594072811608739584170445.
- The last sixteen digits of this number are: 1608739584170445 and in
- scientific notation this number can be written as 9.7e53.
- Find the 100 000 000th squarefree fibonacci number.
- Give as your answer its last sixteen digits followed by a comma followed
- by the number in scientific notation (rounded to one digit after the
- decimal point).
- For the 200th squarefree number the answer would have been:
- 1608739584170445,9.7e53
- Note:
- For this problem, assume that for every prime p, the first fibonacci
- number divisible by p is not divisible by p^2 (this is part of Wall's
- conjecture). This has been verified for primes ≤ 3·10^15, but has not been
- proven in general.
- If it happens that the conjecture is false, then the accepted answer to
- this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci
- number, rather it represents only a lower bound for that number.
- Answer: a0819cfe3be6a04645b8d4fe2345e184
- Problem 400
- ===========
- A Fibonacci tree is a binary tree recursively defined as:
- • T(0) is the empty tree.
- • T(1) is the binary tree with only one node.
- • T(k) consists of a root node that has T(k-1) and T(k-2) as children.
- On such a tree two players play a take-away game. On each turn a player
- selects a node and removes that node along with the subtree rooted at that
- node.
- The player who is forced to take the root node of the entire tree loses.
- Here are the winning moves of the first player on the first turn for T(k)
- from k=1 to k=6.
- Let f(k) be the number of winning moves of the first player (i.e. the
- moves for which the second player has no winning strategy) on the first
- turn of the game when this game is played on T(k).
- For example, f(5) = 1 and f(10) = 17.
- Find f(10000). Give the last 18 digits of your answer.
- Answer: 60aa790c07af1446c1e2deba72543a1f
- Problem 401
- ===========
- The divisors of 6 are 1,2,3 and 6.
- The sum of the squares of these numbers is 1+4+9+36=50.
- Let sigma2(n) represent the sum of the squares of the divisors of n.Thus
- sigma2(6)=50.
- Let SIGMA2 represent the summatory function of sigma2, that is
- SIGMA2(n)=∑sigma2(i) for i=1 to n.
- The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.
- Find SIGMA2(10^15) modulo 10^9.
- Answer: 982a249d8b45ef10c98c32dabac00751
- Problem 402
- ===========
- It can be shown that the polynomial n^4 + 4n^3 + 2n^2 + 5n is a multiple
- of 6 for every integer n. It can also be shown that 6 is the largest
- integer satisfying this property.
- Define M(a, b, c) as the maximum m such that n^4 + an^3 + bn^2 + cn is a
- multiple of m for all integers n. For example, M(4, 2, 5) = 6.
- Also, define S(N) as the sum of M(a, b, c) for all 0 < a, b, c ≤ N.
- We can verify that S(10) = 1972 and S(10000) = 2024258331114.
- Let F[k] be the Fibonacci sequence:
- F[0] = 0, F[1] = 1 and
- F[k] = F[k-1] + F[k-2] for k ≥ 2.
- Find the last 9 digits of Σ S(F[k]) for 2 ≤ k ≤ 1234567890123.
- Answer: fa7ae8e9243f01b0eac10ec5aaff1f42
- Problem 403
- ===========
- For integers a and b, we define D(a, b) as the domain enclosed by the
- parabola y = x^2 and the line y = a·x + b:
- D(a, b) = { (x, y) | x^2 ≤ y ≤ a·x + b }.
- L(a, b) is defined as the number of lattice points contained in D(a, b).
- For example, L(1, 2) = 8 and L(2, -1) = 1.
- We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such
- that the area of D(a, b) is a rational number and |a|,|b| ≤ N.
- We can verify that S(5) = 344 and S(100) = 26709528.
- Find S(10^12). Give your answer mod 10^8.
- Answer: 18224771
- Problem 404
- ===========
- E[a] is an ellipse with an equation of the form x^2 + 4y^2 = 4a^2.
- E[a]' is the rotated image of E[a] by θ degrees counterclockwise around
- the origin O(0, 0) for 0° < θ < 90°.
- b is the distance to the origin of the two intersection points closest to
- the origin and c is the distance of the two other intersection points.
- We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a,
- b and c are positive integers.
- For example, (209, 247, 286) is a canonical ellipsoidal triplet.
- Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b,
- c) for a ≤ N.
- It can be verified that C(10^3) = 7, C(10^4) = 106 and C(10^6) = 11845.
- Find C(10^17).
- p_404_c_ellipse.gif
- Answer: 2d1bc4b93bbc19d9e70c5b04338dea2e
- Problem 405
- ===========
- We wish to tile a rectangle whose length is twice its width.
- Let T(0) be the tiling consisting of a single rectangle.
- For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the
- following manner:
- The following animation demonstrates the tilings T(n) for n from 0 to 5:
- Let f(n) be the number of points where four tiles meet in T(n).
- For example, f(1) = 0, f(4) = 82 and f(10^9) mod 17^7 = 126897180.
- Find f(10^k) for k = 10^18, give your answer modulo 17^7.
- p_405_tile1.png
- p_405_tile2.gif
- Answer: 93b712426b768586f88d0bfe597842e6
- Problem 406
- ===========
- We are trying to find a hidden number selected from the set of integers
- {1, 2, ..., n} by asking questions. Each number (question) we ask, we get
- one of three possible answers:
- • "Your guess is lower than the hidden number" (and you incur a cost of
- a), or
- • "Your guess is higher than the hidden number" (and you incur a cost of
- b), or
- • "Yes, that's it!" (and the game ends).
- Given the value of n, a, and b, an optimal strategy minimizes the total
- cost for the worst possible case.
- For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2"
- as our first question.
- If we are told that 2 is higher than the hidden number (for a cost of
- b=3), then we are sure that "1" is the hidden number (for a total cost of
- 3).
- If we are told that 2 is lower than the hidden number (for a cost of a=2),
- then our next question will be "4".
- If we are told that 4 is higher than the hidden number (for a cost of
- b=3), then we are sure that "3" is the hidden number (for a total cost of
- 2+3=5).
- If we are told that 4 is lower than the hidden number (for a cost of a=2),
- then we are sure that "5" is the hidden number (for a total cost of
- 2+2=4).
- Thus, the worst-case cost achieved by this strategy is 5. It can also be
- shown that this is the lowest worst-case cost that can be achieved. So, in
- fact, we have just described an optimal strategy for the given values of
- n, a, and b.
- Let C(n, a, b) be the worst-case cost achieved by an optimal strategy for
- the given values of n, a, and b.
- Here are a few examples:
- C(5, 2, 3) = 5
- C(500, √2, √3) = 13.22073197...
- C(20000, 5, 7) = 82
- C(2000000, √5, √7) = 49.63755955...
- Let F[k] be the Fibonacci numbers: F[k] = F[k-1] + F[k-2] with base cases
- F[1] = F[2] = 1.
- Find ∑[1≤k≤30] C(10^12, √k, √F[k]), and give your answer rounded to 8
- decimal places behind the decimal point.
- Answer: 0766b1ee975f5674d30fd6c3c934c6e0
- Problem 407
- ===========
- If we calculate a^2 mod 6 for 0 ≤ a ≤ 5 we get: 0,1,4,3,4,1.
- The largest value of a such that a^2 ≡ a mod 6 is 4.
- Let's call M(n) the largest value of a < n such that a^2 ≡ a (mod n).
- So M(6) = 4.
- Find ∑M(n) for 1 ≤ n ≤ 10^7.
- Answer: f4da34a4b357123cb142739a52e010f2
- Problem 408
- ===========
- Let's call a lattice point (x, y) inadmissible if x, y and x + y are all
- positive perfect squares.
- For example, (9, 16) is inadmissible, while (0, 4), (3, 1) and (9, 4) are
- not.
- Consider a path from point (x[1], y[1]) to point (x[2], y[2]) using only
- unit steps north or east.
- Let's call such a path admissible if none of its intermediate points are
- inadmissible.
- Let P(n) be the number of admissible paths from (0, 0) to (n, n).
- It can be verified that P(5) = 252, P(16) = 596994440 and P(1000) mod
- 1 000 000 007 = 341920854.
- Find P(10 000 000) mod 1 000 000 007.
- Answer: 2c09e247c6144c16cae2358d316affd9
- Problem 409
- ===========
- Let n be a positive integer. Consider nim positions where:
- • There are n non-empty piles.
- • Each pile has size less than 2^n.
- • No two piles have the same size.
- Let W(n) be the number of winning nim positions satisfying the
- aboveconditions (a position is winning if the first player has a winning
- strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360
- and W(100) mod 1 000 000 007 = 384777056.
- Find W(10 000 000) mod 1 000 000 007.
- Answer: 56c32e75a2656ec08ce177089bda2f53
- Problem 410
- ===========
- Let C be the circle with radius r, x^2 + y^2 = r^2. We choose two points
- P(a, b) and Q(-a, c) so that the line passing through P and Q is tangent
- to C.
- For example, the quadruplet (r, a, b, c) = (2, 6, 2, -7) satisfies this
- property.
- Let F(R, X) be the number of the integer quadruplets (r, a, b, c) with
- this property, and with 0 < r ≤ R and 0 < a ≤ X.
- We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384.
- Find F(10^8, 10^9) + F(10^9, 10^8).
- Answer: 45826f3a23aa321f97acb1d2a8f2170b
- Problem 411
- ===========
- Let n be a positive integer. Suppose there are stations at the coordinates
- (x, y) = (2^i mod n, 3^i mod n) for 0 ≤ i ≤ 2n. We will consider stations
- with the same coordinates as the same station.
- We wish to form a path from (0, 0) to (n, n) such that the x and y
- coordinates never decrease.
- Let S(n) be the maximum number of stations such a path can pass through.
- For example, if n = 22, there are 11 distinct stations, and a valid path
- can pass through at most 5 stations. Therefore, S(22) = 5.The case is
- illustrated below, with an example of an optimal path:
- It can also be verified that S(123) = 14 and S(10000) = 48.
- Find ∑ S(k^5) for 1 ≤ k ≤ 30.
- Answer: 9936352
- Problem 412
- ===========
- For integers m, n (0 ≤ n < m), let L(m, n) be an m×m grid with the
- top-right n×n grid removed.
- For example, L(5, 3) looks like this:
- We want to number each cell of L(m, n) with consecutive integers 1, 2, 3,
- ... such that the number in every cell is smaller than the number below it
- and to the left of it.
- For example, here are two valid numberings of L(5, 3):
- Let LC(m, n) be the number of valid numberings of L(m, n).
- It can be verified that LC(3, 0) = 42, LC(5, 3) = 250250, LC(6, 3) =
- 406029023400 and LC(10, 5) mod 76543217 = 61251715.
- Find LC(10000, 5000) mod 76543217.
- Answer: 38788800
- Problem 413
- ===========
- We say that a d-digit positive number (no leading zeros) is a one-child
- number if exactly one of its sub-strings is divisible by d.
- For example, 5671 is a 4-digit one-child number. Among all its sub-strings
- 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4.
- Similarly, 104 is a 3-digit one-child number because only 0 is divisible
- by 3.
- 1132451 is a 7-digit one-child number because only 245 is divisible by 7.
- Let F(N) be the number of the one-child numbers less than N.
- We can verify that F(10) = 9, F(10^3) = 389 and F(10^7) = 277674.
- Find F(10^19).
- Answer: 569ad33af889215704df5a9e278aa004
- Problem 414
- ===========
- 6174 is a remarkable number; if we sort its digits in increasing order and
- subtract that number from the number you get when you sort the digits in
- decreasing order, we get 7641-1467=6174.
- Even more remarkable is that if we start from any 4 digit number and
- repeat this process of sorting and subtracting, we'll eventually end up
- with 6174 or immediately with 0 if all digits are equal.
- This also works with numbers that have less than 4 digits if we pad the
- number with leading zeroes until we have 4 digits.
- E.g. let's start with the number 0837:
- 8730-0378=8352
- 8532-2358=6174
- 6174 is called the Kaprekar constant. The process of sorting and
- subtracting and repeating this until either 0 or the Kaprekar constant is
- reached is called the Kaprekar routine.
- We can consider the Kaprekar routine for other bases and number of digits.
- Unfortunately, it is not guaranteed a Kaprekar constant exists in all
- cases; either the routine can end up in a cycle for some input numbers or
- the constant the routine arrives at can be different for different input
- numbers.
- However, it can be shown that for 5 digits and a base b = 6t+3≠9, a
- Kaprekar constant exists.
- E.g. base 15: (10,4,14,9,5)[15]
- base 21: (14,6,20,13,7)[21]
- Define C[b] to be the Kaprekar constant in base b for 5 digits.Define the
- function sb(i) to be
- • 0 if i = C[b] or if i written in base b consists of 5 identical digits
- • the number of iterations it takes the Kaprekar routine in base b to
- arrive at C[b], otherwise
- Note that we can define sb(i) for all integers i < b^5. If i written in
- base b takes less than 5 digits, the number is padded with leading zero
- digits until we have 5 digits before applying the Kaprekar routine.
- Define S(b) as the sum of sb(i) for 0 < i < b^5.
- E.g. S(15) = 5274369
- S(111) = 400668930299
- Find the sum of S(6k+3) for 2 ≤ k ≤ 300.
- Give the last 18 digits as your answer.
- Answer: 42f095bdfd71e1ae4ae0ceead1bb1802
- Problem 415
- ===========
- A set of lattice points S is called a titanic set if there exists a line
- passing through exactly two points in S.
- An example of a titanic set is S = {(0, 0), (0, 1), (0, 2), (1, 1), (2,
- 0), (1, 0)}, where the line passing through (0, 1) and (2, 0) does not
- pass through any other point in S.
- On the other hand, the set {(0, 0), (1, 1), (2, 2), (4, 4)} is not a
- titanic set since the line passing through any two points in the set also
- passes through the other two.
- For any positive integer N, let T(N) be the number of titanic sets S whose
- every point (x, y) satisfies 0 ≤ x, y ≤ N.It can be verified that T(1) =
- 11, T(2) = 494, T(4) = 33554178, T(111) mod 10^8 = 13500401 and
- T(10^5) mod 10^8 = 63259062.
- Find T(10^11) mod 10^8.
- Answer: 55859742
- Problem 416
- ===========
- A row of n squares contains a frog in the leftmost square. By successive
- jumps the frog goes to the rightmost square and then back to the leftmost
- square. On the outward trip he jumps one, two or three squares to the
- right, and on the homeward trip he jumps to the left in a similar manner.
- He cannot jump outside the squares. He repeats the round-trip travel m
- times.
- Let F(m, n) be the number of the ways the frog can travel so that at most
- one square remains unvisited.
- For example, F(1, 3) = 4, F(1, 4) = 15, F(1, 5) = 46, F(2, 3) = 16 and
- F(2, 100) mod 10^9 = 429619151.
- Find the last 9 digits of F(10, 10^12).
- Answer: 6f398386fdfec57ac166d4970c2bcad2
- Problem 417
- ===========
- A unit fraction contains 1 in the numerator. The decimal representation of
- the unit fractions with denominators 2 to 10 are given:
- 1/2 = 0.5
- 1/3 = 0.(3)
- 1/4 = 0.25
- 1/5 = 0.2
- 1/6 = 0.1(6)
- 1/7 = 0.(142857)
- 1/8 = 0.125
- 1/9 = 0.(1)
- 1/10 = 0.1
- Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can
- be seen that 1/7 has a 6-digit recurring cycle.
- Unit fractions whose denominator has no other prime factors than 2 and/or
- 5 are not considered to have a recurring cycle.
- We define the length of the recurring cycle of those unit fractions as 0.
- Let L(n) denote the length of the recurring cycle of 1/n.You are given
- that ∑L(n) for 3 ≤ n ≤ 1 000 000 equals 55535191115.
- Find ∑L(n) for 3 ≤ n ≤ 100 000 000
- Answer: 93a7df08c972f1e7788516d056a7e016
- Problem 418
- ===========
- Let n be a positive integer. An integer triple (a, b, c) is called a
- factorisation triple of n if:
- • 1 ≤ a ≤ b ≤ c
- • a·b·c = n.
- Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n
- which minimises c / a. One can show that this triple is unique.
- For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872.
- Find f(43!).
- Answer: b032468ddb4847d8a2273789379753f5
- Problem 419
- ===========
- The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221,
- 1113213211, ...
- The sequence starts with 1 and all other members are obtained by
- describing the previous member in terms of consecutive digits.
- It helps to do this out loud:
- 1 is 'one one' → 11
- 11 is 'two ones' → 21
- 21 is 'one two and one one' → 1211
- 1211 is 'one one, one two and two ones' → 111221
- 111221 is 'three ones, two twos and one one' → 312211
- ...
- Define A(n), B(n) and C(n) as the number of ones, twos and threes in the
- n'th element of the sequence respectively.
- One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625.
- Find A(n), B(n) and C(n) for n = 10^12.
- Give your answer modulo 2^30 and separate your values for A, B and C by a
- comma.
- E.g. for n = 40 the answer would be 31254,20259,11625
- Answer: b27db655498b3d64ad4338fcdc9d178f
- Problem 420
- ===========
- A positive integer matrix is a matrix whose elements are all positive
- integers.
- Some positive integer matrices can be expressed as a square of a positive
- integer matrix in two different ways. Here is an example:
- We define F(N) as the number of the 2x2 positive integer matrices which
- have a trace less than N and which can be expressed as a square of a
- positive integer matrix in two different ways.
- We can verify that F(50) = 7 and F(1000) = 1019.
- Find F(10^7).
- p_420_matrix.gif
- Answer: 145159332
- Problem 421
- ===========
- Numbers of the form n^15+1 are composite for every integer n > 1.
- For positive integers n and m let s(n,m) be defined as the sum of the
- distinct prime factors of n^15+1 not exceeding m.
- E.g. 2^15+1 = 3×3×11×331.
- So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345.
- Also 10^15+1 = 7×11×13×211×241×2161×9091.
- So s(10,100) = 31 and s(10,1000) = 483.
- Find ∑ s(n,10^8) for 1 ≤ n ≤ 10^11.
- Answer: 481fcc5ff16ccf1645fb136c123ed660
- Problem 422
- ===========
- Let H be the hyperbola defined by the equation 12x^2 + 7xy - 12y^2 = 625.
- Next, define X as the point (7, 1). It can be seen that X is in H.
- Now we define a sequence of points in H, {P[i] : i ≥ 1}, as:
- • P[1] = (13, 61/4).
- • P[2] = (-43/6, -4).
- • For i > 2, P[i] is the unique point in H that is different from P[i-1]
- and such that line P[i]P[i-1] is parallel to line P[i-2]X. It can be
- shown that P[i] is well-defined, and that its coordinates are always
- rational.
- You are given that P[3] = (-19/2, -229/24), P[4] = (1267/144, -37/12) and
- P[7] = (17194218091/143327232, 274748766781/1719926784).
- Find P[n] for n = 11^14 in the following format:
- If P[n] = (a/b, c/d) where the fractions are in lowest terms and the
- denominators are positive, then the answer is (a + b + c + d) mod
- 1 000 000 007.
- For n = 7, the answer would have been: 806236837.
- Answer: 92060460
- Problem 423
- ===========
- Let n be a positive integer.
- A 6-sided die is thrown n times. Let c be the number of pairs of
- consecutive throws that give the same value.
- For example, if n = 7 and the values of the die throws are
- (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the
- same value:
- (1,1,5,6,6,6,3)
- (1,1,5,6,6,6,3)
- (1,1,5,6,6,6,3)
- Therefore, c = 3 for (1,1,5,6,6,6,3).
- Define C(n) as the number of outcomes of throwing a 6-sided die n times
- such that c does not exceed π(n).^1
- For example, C(3) = 216, C(4) = 1290, C(11) = 361912500 and C(24) =
- 4727547363281250000.
- Define S(L) as ∑ C(n) for 1 ≤ n ≤ L.
- For example, S(50) mod 1 000 000 007 = 832833871.
- Find S(50 000 000) mod 1 000 000 007.
- ^1 π denotes the prime-counting function, i.e. π(n) is the number of
- primes ≤ n.
- Answer: e2add9d46ebd8ba59a07dca791cd629b
- Problem 424
- ===========
- The above is an example of a cryptic kakuro (also known as cross sums, or
- even sums cross) puzzle, with its final solution on the right. (The common
- rules of kakuro puzzles can be found easily on numerous internet sites.
- Other related information can also be currently found at [1]krazydad.com
- whose author has provided the puzzle data for this challenge.)
- The downloadable text file ([2]kakuro200.txt) contains the description of
- 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file
- is the above example which is coded as follows:
- 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
- The first character is a numerical digit indicating the size of the
- information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7
- (for a 6x6 puzzle) followed by a comma (,). The extra top line and left
- column are needed to insert information.
- The content of each cell is then described and followed by a comma, going
- left to right and starting with the top line.
- X = Gray cell, not required to be filled by a digit.
- O (upper case letter)= White empty cell to be filled by a digit.
- A = Or any one of the upper case letters from A to J to be replaced by its
- equivalent digit in the solved puzzle.
- ( ) = Location of the encrypted sums. Horizontal sums are preceded by a
- lower case "h" and vertical sums are preceded by a lower case "v". Those
- are followed by one or two upper case letters depending if the sum is a
- single digit or double digit one. For double digit sums, the first letter
- would be for the "tens" and the second one for the "units". When the cell
- must contain information for both a horizontal and a vertical sum, the
- first one is always for the horizontal sum and the two are separated by a
- comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets
- is also immediately followed by a comma.
- The description of the last cell is followed by a Carriage Return/Line
- Feed (CRLF) instead of a comma.
- The required answer to each puzzle is based on the value of each letter
- necessary to arrive at the solution and according to the alphabetical
- order. As indicated under the example puzzle, its answer would be
- 8426039571. At least 9 out of the 10 encrypting letters are always part of
- the problem description. When only 9 are given, the missing one must be
- assigned the remaining digit.
- You are given that the sum of the answers for the first 10 puzzles in the
- file is 64414157580.
- Find the sum of the answers for the 200 puzzles.
- Visible links
- 1. http://krazydad.com/
- 2. kakuro200.txt
- Answer: 1059760019628
- Problem 425
- ===========
- Two positive numbers A and B are said to be connected (denoted by "A ↔ B")
- if one of these conditions holds:
- (1) A and B have the same length and differ in exactly one digit; for
- example, 123 ↔ 173.
- (2) Adding one digit to the left of A (or B) makes B (or A); for example,
- 23 ↔ 223 and 123 ↔ 23.
- We call a prime P a 2's relative if there exists a chain of connected
- primes between 2 and P and no prime in the chain exceeds P.
- For example, 127 is a 2's relative. One of the possible chains is shown
- below:
- 2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127
- However, 11 and 103 are not 2's relatives.
- Let F(N) be the sum of the primes ≤ N which are not 2's relatives.
- We can verify that F(10^3) = 431 and F(10^4) = 78728.
- Find F(10^7).
- Answer: 3d229894ba4c585138125e802af2d06e
- Problem 426
- ===========
- Consider an infinite row of boxes. Some of the boxes contain a ball. For
- example, an initial configuration of 2 consecutive occupied boxes followed
- by 2 empty boxes, 2 occupied boxes, 1 empty box, and 2 occupied boxes can
- be denoted by the sequence (2, 2, 2, 1, 2), in which the number of
- consecutive occupied and empty boxes appear alternately.
- A turn consists of moving each ball exactly once according to the
- following rule: Transfer the leftmost ball which has not been moved to the
- nearest empty box to its right.
- After one turn the sequence (2, 2, 2, 1, 2) becomes (2, 2, 1, 2, 3) as can
- be seen below; note that we begin the new sequence starting at the first
- occupied box.
- A system like this is called a Box-Ball System or BBS for short.
- It can be shown that after a sufficient number of turns, the system
- evolves to a state where the consecutive numbers of occupied boxes is
- invariant. In the example below, the consecutive numbers of occupied boxes
- evolves to [1, 2, 3]; we shall call this the final state.
- We define the sequence {t[i]}:
- • s[0] = 290797
- • s[k+1] = s[k]^2 mod 50515093
- • t[k] = (s[k] mod 64) + 1
- Starting from the initial configuration (t[0], t[1], …, t[10]), the final
- state becomes [1, 3, 10, 24, 51, 75].
- Starting from the initial configuration (t[0], t[1], …, t[10 000 000]),
- find the final state.
- Give as your answer the sum of the squares of the elements of the final
- state. For example, if the final state is [1, 2, 3] then 14 ( = 1^2 + 2^2
- + 3^2) is your answer.
- p_426_baxball1.gif
- p_426_baxball2.gif
- Answer: b5d8157a351482da47da0512ca374007
- Problem 427
- ===========
- A sequence of integers S = {s[i]} is called an n-sequence if it has n
- elements and each element s[i] satisfies 1 ≤ s[i] ≤ n. Thus there are n^n
- distinct n-sequences in total.For example, the sequence S = {1, 5, 5, 10,
- 7, 7, 7, 2, 3, 7} is a 10-sequence.
- For any sequence S, let L(S) be the length of the longest contiguous
- subsequence of S with the same value.For example, for the given sequence S
- above, L(S) = 3, because of the three consecutive 7's.
- Let f(n) = ∑ L(S) for all n-sequences S.
- For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121.
- Find f(7 500 000) mod 1 000 000 009.
- Answer: 97138867
- Problem 428
- ===========
- Let a, b and c be positive numbers.
- Let W, X, Y, Z be four collinear points where |WX| = a, |XY| = b, |YZ| = c
- and |WZ| = a + b + c.
- Let C[in] be the circle having the diameter XY.
- Let C[out] be the circle having the diameter WZ.
- The triplet (a, b, c) is called a necklace triplet if you can place k ≥ 3
- distinct circles C[1], C[2], ..., C[k] such that:
- • C[i] has no common interior points with any C[j] for 1 ≤ i, j ≤ k and
- i ≠ j,
- • C[i] is tangent to both C[in] and C[out] for 1 ≤ i ≤ k,
- • C[i] is tangent to C[i+1] for 1 ≤ i < k, and
- • C[k] is tangent to C[1].
- For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can
- be shown that (2, 2, 5) is not.
- Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c
- are positive integers, and b ≤ n.For example, T(1) = 9, T(20) = 732 and
- T(3000) = 438106.
- Find T(1 000 000 000).
- Answer: c6010c109b66b34bf3594e63eb58b446
- Problem 429
- ===========
- A unitary divisor d of a number n is a divisor of n that has the property
- gcd(d, n/d) = 1.
- The unitary divisors of 4! = 24 are 1, 3, 8 and 24.
- The sum of their squares is 1^2 + 3^2 + 8^2 + 24^2 = 650.
- Let S(n) represent the sum of the squares of the unitary divisors of n.
- Thus S(4!)=650.
- Find S(100 000 000!) modulo 1 000 000 009.
- Answer: 98792821
- Problem 430
- ===========
- N disks are placed in a row, indexed 1 to N from left to right.
- Each disk has a black side and white side. Initially all disks show their
- white side.
- At each turn, two, not necessarily distinct, integers A and B between 1
- and N (inclusive) are chosen uniformly at random.
- All disks with an index from A to B (inclusive) are flipped.
- The following example shows the case N = 8. At the first turn A = 5 and B
- = 2, and at the second turn A = 4 and B = 6.
- Let E(N, M) be the expected number of disks that show their white side
- after M turns.
- We can verify that E(3, 1) = 10/9, E(3, 2) = 5/3, E(10, 4) ≈ 5.157 and
- E(100, 10) ≈ 51.893.
- Find E(10^10, 4000).
- Give your answer rounded to 2 decimal places behind the decimal point.
- p_430_flips.gif
- Answer: 32b0825d7a110a1a220e80629c413411
- Problem 431
- ===========
- Fred the farmer arranges to have a new storage silo installed on his farm
- and having an obsession for all things square he is absolutely devastated
- when he discovers that it is circular. Quentin, the representative from
- the company that installed the silo, explains that they only manufacture
- cylindrical silos, but he points out that it is resting on a square base.
- Fred is not amused and insists that it is removed from his property.
- Quick thinking Quentin explains that when granular materials are delivered
- from above a conical slope is formed and the natural angle made with the
- horizontal is called the angle of repose. For example if the angle of
- repose, $\alpha = 30$ degrees, and grain is delivered at the centre of the
- silo then a perfect cone will form towards the top of the cylinder. In the
- case of this silo, which has a diameter of 6m, the amount of space wasted
- would be approximately 32.648388556 m^3. However, if grain is delivered at
- a point on the top which has a horizontal distance of $x$ metres from the
- centre then a cone with a strangely curved and sloping base is formed. He
- shows Fred a picture.
- We shall let the amount of space wasted in cubic metres be given by
- $V(x)$. If $x = 1.114785284$, which happens to have three squared decimal
- places, then the amount of space wasted, $V(1.114785284) \approx 36$.
- Given the range of possible solutions to this problem there is exactly one
- other option: $V(2.511167869) \approx 49$. It would be like knowing that
- the square is king of the silo, sitting in splendid glory on top of your
- grain.
- Fred's eyes light up with delight at this elegant resolution, but on
- closer inspection of Quentin's drawings and calculations his happiness
- turns to despondency once more. Fred points out to Quentin that it's the
- radius of the silo that is 6 metres, not the diameter, and the angle of
- repose for his grain is 40 degrees. However, if Quentin can find a set of
- solutions for this particular silo then he will be more than happy to keep
- it.
- If Quick thinking Quentin is to satisfy frustratingly fussy Fred the
- farmer's appetite for all things square then determine the values of $x$
- for all possible square space wastage options and calculate $\sum x$
- correct to 9 decimal places.
- p_431_grain_silo.png
- Answer: 5e5d81aa8bfaf92f68cdef0154c5c238
- Problem 432
- ===========
- Let S(n,m) = ∑φ(n × i) for 1 ≤ i ≤ m. (φ is Euler's totient function)
- You are given that S(510510,10^6 )= 45480596821125120.
- Find S(510510,10^11).
- Give the last 9 digits of your answer.
- Answer: e171c2872d650e47589842faa80f5707
- Problem 433
- ===========
- Let E(x[0], y[0]) be the number of steps it takes to determine the
- greatest common divisor of x[0] and y[0] with Euclid's algorithm. More
- formally:
- x[1] = y[0], y[1] = x[0] mod y[0]
- x[n] = y[n-1], y[n] = x[n-1] mod y[n-1]
- E(x[0], y[0]) is the smallest n such that y[n] = 0.
- We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4.
- Define S(N) as the sum of E(x,y) for 1 ≤ x,y ≤ N.
- We have S(1) = 1, S(10) = 221 and S(100) = 39826.
- Find S(5·10^6).
- Answer: 0eeca9fa5cf25a2bfae01f1f04d6cd35
- Problem 434
- ===========
- Recall that a graph is a collection of vertices and edges connecting the
- vertices, and that two vertices connected by an edge are called adjacent.
- Graphs can be embedded in Euclidean space by associating each vertex with
- a point in the Euclidean space.
- A flexible graph is an embedding of a graph where it is possible to move
- one or more vertices continuously so that the distance between at least
- two nonadjacent vertices is altered while the distances between each pair
- of adjacent vertices is kept constant.
- A rigid graph is an embedding of a graph which is not flexible.
- Informally, a graph is rigid if by replacing the vertices with fully
- rotating hinges and the edges with rods that are unbending and inelastic,
- no parts of the graph can be moved independently from the rest of the
- graph.
- The grid graphs embedded in the Euclidean plane are not rigid, as the
- following animation demonstrates:
- However, one can make them rigid by adding diagonal edges to the cells.
- For example, for the 2x3 grid graph, there are 19 ways to make the graph
- rigid:
- Note that for the purposes of this problem, we do not consider changing
- the orientation of a diagonal edge or adding both diagonal edges to a cell
- as a different way of making a grid graph rigid.
- Let R(m,n) be the number of ways to make the m × n grid graph rigid.
- E.g. R(2,3) = 19 and R(5,5) = 23679901
- Define S(N) as ∑R(i,j) for 1 ≤ i, j ≤ N.
- E.g. S(5) = 25021721.
- Find S(100), give your answer modulo 1000000033
- Answer: f51d9fd41a8ce217682321a020be6fec
- Problem 435
- ===========
- The Fibonacci numbers {f[n], n ≥ 0} are defined recursively as f[n] =
- f[n-1] + f[n-2] with base cases f[0] = 0 and f[1] = 1.
- Define the polynomials {F[n], n ≥ 0} as F[n](x) = ∑f[i]x^i for 0 ≤ i ≤ n.
- For example, F[7](x) = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7, and
- F[7](11) = 268357683.
- Let n = 10^15. Find the sum [∑[0≤x≤100] F[n](x)] mod 1307674368000 (=
- 15!).
- Answer: 0f08231a97e872f565a085de75743a1c
- Problem 436
- ===========
- Julie proposes the following wager to her sister Louise.
- She suggests they play a game of chance to determine who will wash the
- dishes.
- For this game, they shall use a generator of independent random numbers
- uniformly distributed between 0 and 1.
- The game starts with S = 0.
- The first player, Louise, adds to S different random numbers from the
- generator until S > 1 and records her last random number 'x'.
- The second player, Julie, continues adding to S different random numbers
- from the generator until S > 2 and records her last random number 'y'.
- The player with the highest number wins and the loser washes the dishes,
- i.e. if y > x the second player wins.
- For example, if the first player draws 0.62 and 0.44, the first player
- turn ends since 0.62+0.44 > 1 and x = 0.44.
- If the second players draws 0.1, 0.27 and 0.91, the second player turn
- ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91.Since y > x, the
- second player wins.
- Louise thinks about it for a second, and objects: "That's not fair".
- What is the probability that the second player wins?
- Give your answer rounded to 10 places behind the decimal point in the form
- 0.abcdefghij
- Answer: d797ed72189f045e8ea48aa960fec1f3
- Problem 437
- ===========
- When we calculate 8^n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3,
- 2, 5, 7.
- As we see all possible values from 1 to 10 occur. So 8 is a primitive root
- of 11.
- But there is more:
- If we take a closer look we see:
- 1+8=9
- 8+9=17≡6 mod 11
- 9+6=15≡4 mod 11
- 6+4=10
- 4+10=14≡3 mod 11
- 10+3=13≡2 mod 11
- 3+2=5
- 2+5=7
- 5+7=12≡1 mod 11.
- So the powers of 8 mod 11 are cyclic with period 10, and 8^n + 8^n+1 ≡
- 8^n+2 (mod 11).
- 8 is called a Fibonacci primitive root of 11.
- Not every prime has a Fibonacci primitive root.
- There are 323 primes less than 10000 with one or more Fibonacci primitive
- roots and the sum of these primes is 1480491.
- Find the sum of the primes less than 100,000,000 with at least one
- Fibonacci primitive root.
- Answer: 74204709657207
- Problem 438
- ===========
- For an n-tuple of integers t = (a[1], ..., a[n]), let (x[1], ..., x[n]) be
- the solutions of the polynomial equation x^n + a[1]x^n-1 + a[2]x^n-2 + ...
- + a[n-1]x + a[n] = 0.
- Consider the following two conditions:
- • x[1], ..., x[n] are all real.
- • If x[1], ..., x[n] are sorted, ⌊x[i]⌋ = i for 1 ≤ i ≤ n. (⌊·⌋: floor
- function.)
- In the case of n = 4, there are 12 n-tuples of integers which satisfy both
- conditions.
- We define S(t) as the sum of the absolute values of the integers in t.
- For n = 4 we can verify that ∑S(t) = 2087 for all n-tuples t which satisfy
- both conditions.
- Find ∑S(t) for n = 7.
- Answer: ?
- Problem 439
- ===========
- Let d(k) be the sum of all divisors of k.
- We define the function S(N) = ∑[1≤i≤N] ∑[1≤j≤N] d(i·j).
- For example, S(3) = d(1) + d(2) + d(3) + d(2) + d(4) + d(6) + d(3) + d(6)
- + d(9) = 59.
- You are given that S(10^3) = 563576517282 and S(10^5) mod 10^9 =
- 215766508.
- Find S(10^11) mod 10^9.
- Answer: ?
- Problem 440
- ===========
- We want to tile a board of length n and height 1 completely, with either 1
- × 2 blocks or 1 × 1 blocks with a single decimal digit on top:
- For example, here are some of the ways to tile a board of length n = 8:
- Let T(n) be the number of ways to tile a board of length n as described
- above.
- For example, T(1) = 10 and T(2) = 101.
- Let S(L) be the triple sum ∑[a,b,c] gcd(T(c^a), T(c^b)) for 1 ≤ a, b, c ≤
- L.
- For example:
- S(2) = 10444
- S(3) = 1292115238446807016106539989
- S(4) mod 987 898 789 = 670616280.
- Find S(2000) mod 987 898 789.
- Answer: ?
- Problem 441
- ===========
- For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer
- pairs p and q which satisfy all of these conditions:
- • 1 ≤ p < q ≤ M
- • p + q ≥ M
- • p and q are coprime.
- We also define S(N) as the sum of R(i) for 2 ≤ i ≤ N.
- We can verify that S(2) = R(2) = 1/2, S(10) ≈ 6.9147 and S(100) ≈ 58.2962.
- Find S(10^7). Give your answer rounded to four decimal places.
- Answer: 152cc265f5461c5055db95a122280416
- Problem 442
- ===========
- An integer is called eleven-free if its decimal expansion does not contain
- any substring representing a power of 11 except 1.
- For example, 2404 and 13431 are eleven-free, while 911 and 4121331 are
- not.
- Let E(n) be the nth positive eleven-free integer. For example, E(3) = 3,
- E(200) = 213 and E(500 000) = 531563.
- Find E(10^18).
- Answer: c31bb13db787bce9a169dce600aec863
- Problem 443
- ===========
- Let g(n) be a sequence defined as follows:
- g(4) = 13,
- g(n) = g(n-1) + gcd(n, g(n-1)) for n > 4.
- The first few values are:
- n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
- g(n) 13 14 16 17 18 27 28 29 30 31 32 33 34 51 54 55 60 ...
- You are given that g(1 000) = 2524 and g(1 000 000) = 2624152.
- Find g(10^15).
- Answer: 28f9d9a9bf8fb3d606e0b711b59f42aa
- Problem 444
- ===========
- A group of p people decide to sit down at a round table and play a
- lottery-ticket trading game. Each person starts off with a
- randomly-assigned, unscratched lottery ticket. Each ticket, when
- scratched, reveals a whole-pound prize ranging anywhere from £1 to £p,
- with no two tickets alike. The goal of the game is for each person to
- maximize his ticket winnings upon leaving the game.
- An arbitrary person is chosen to be the first player. Going around the
- table, each player has only one of two options:
- 1. The player can scratch his ticket and reveal its worth to everyone at
- the table.
- 2. The player can trade his unscratched ticket for a previous player's
- scratched ticket, and then leave the game with that ticket. The previous
- player then scratches his newly-acquired ticket and reveals its worth to
- everyone at the table.
- The game ends once all tickets have been scratched. All players still
- remaining at the table must leave with their currently-held tickets.
- Assume that each player uses the optimal strategy for maximizing the
- expected value of his ticket winnings.
- Let E(p) represent the expected number of players left at the table when
- the game ends in a game consisting of p players (e.g. E(111) = 5.2912 when
- rounded to 5 significant digits).
- Let S[1](N) = E(p)
- Let S[k](N) = S[k-1](p) for k > 1
- Find S[20](10^14) and write the answer in scientific notation rounded to
- 10 significant digits. Use a lowercase e to separate mantissa and exponent
- (e.g. S[3](100) = 5.983679014e5).
- Answer: e6745c386ba3c0de1bf56897e453c7c8
- Problem 445
- ===========
- For every integer n>1, the family of functions f[n,a,b] is defined by
- f[n,a,b](x)≡ax+b mod n for a,b,x integer and 0<a<n, 0≤b<n, 0≤x<n.
- We will call f[n,a,b] a retraction if f[n,a,b](f[n,a,b](x))≡f[n,a,b](x)
- mod n for every 0≤x<n.
- Let R(n) be the number of retractions for n.
- You are given that
- ∑ R(c) for c=C(100 000,k), and 1 ≤ k ≤99 999 ≡628701600 (mod 1 000 000
- 007).
- (C(n,k) is the binomial coefficient).
- Find ∑ R(c) for c=C(10 000 000,k), and 1 ≤k≤ 9 999 999.
- Give your answer modulo 1 000 000 007.
- Answer: ?
- Problem 446
- ===========
- For every integer n>1, the family of functions f[n,a,b] is defined by
- f[n,a,b](x)≡ax+b mod n for a,b,x integer and 0<a<n, 0≤b<n, 0≤x<n.
- We will call f[n,a,b] a retraction if f[n,a,b](f[n,a,b](x))≡f[n,a,b](x)
- mod n for every 0≤x<n.
- Let R(n) be the number of retractions for n.
- F(N)=∑R(n^4+4) for 1≤n≤N.
- F(1024)=77532377300600.
- Find F(10^7) (mod 1 000 000 007)
- Answer: ?
- Problem 447
- ===========
- For every integer n>1, the family of functions f[n,a,b] is defined by
- f[n,a,b](x)≡ax+b mod n for a,b,x integer and 0<a<n, 0≤b<n, 0≤x<n.
- We will call f[n,a,b] a retraction if f[n,a,b](f[n,a,b](x))≡f[n,a,b](x)
- mod n for every 0≤x<n.
- Let R(n) be the number of retractions for n.
- F(N)=∑R(n) for 2≤n≤N.
- F(10^7)≡638042271 (mod 1 000 000 007).
- Find F(10^14) (mod 1 000 000 007).
- Answer: ?
- Problem 448
- ===========
- The function lcm(a,b) denotes the least common multiple of a and b.
- Let A(n) be the average of the values of lcm(n,i) for 1≤i≤n.
- E.g: A(2)=(2+2)/2=2 and A(10)=(10+10+30+20+10+30+70+40+90+10)/10=32.
- Let S(n)=∑A(k) for 1≤k≤n.
- S(100)=122726.
- Find S(99999999019) mod 999999017.
- Answer: 106467648
- Problem 449
- ===========
- Phil the confectioner is making a new batch of chocolate covered candy.
- Each candy centre is shaped like an ellipsoid of revolution defined by the
- equation: b^2x^2 + b^2y^2 + a^2z^2 = a^2b^2.
- Phil wants to know how much chocolate is needed to cover one candy centre
- with a uniform coat of chocolate one millimeter thick.
- If a=1 mm and b=1 mm, the amount of chocolate required is 28 π mm^3
- 3
- If a=2 mm and b=1 mm, the amount of chocolate required is approximately
- 60.35475635 mm^3.
- Find the amount of chocolate in mm^3 required if a=3 mm and b=1 mm. Give
- your answer as the number rounded to 8 decimal places behind the decimal
- point.
- Answer: 8ac19d0d06980691526883bc8c0950ef
- Problem 450
- ===========
- A hypocycloid is the curve drawn by a point on a small circle rolling
- inside a larger circle. The parametric equations of a hypocycloid centered
- at the origin, and starting at the right most point is given by:
- $x(t) = (R - r) \cos(t) + r \cos(\frac {R - r} r t)$
- $y(t) = (R - r) \sin(t) - r \sin(\frac {R - r} r t)$
- Where R is the radius of the large circle and r the radius of the small
- circle.
- Let $C(R, r)$ be the set of distinct points with integer coordinates on
- the hypocycloid with radius R and r and for which there is a corresponding
- value of t such that $\sin(t)$ and $\cos(t)$ are rational numbers.
- Let $S(R, r) = \sum_{(x,y) \in C(R, r)} |x| + |y|$ be the sum of the
- absolute values of the x and y coordinates of the points in $C(R, r)$.
- Let $T(N) = \sum_{R = 3}^N \sum_{r=1}^{\lfloor \frac {R - 1} 2 \rfloor}
- S(R, r)$ be the sum of $S(R, r)$ for R and r positive integers, $R\leq N$
- and $2r < R$.
- You are given:
- C(3, 1) = {(3, 0), (-1, 2), (-1,0), (-1,-2)}
- C(2500, 1000) =
- {(2500, 0), (772, 2376), (772, -2376), (516, 1792), (516, -1792), (500,
- 0), (68, 504), (68, -504),
- (-1356, 1088), (-1356, -1088), (-1500, 1000), (-1500, -1000)}
- Note: (-625, 0) is not an element of C(2500, 1000) because $\sin(t)$ is
- not a rational number for the corresponding values of t.
- S(3, 1) = (|3| + |0|) + (|-1| + |2|) + (|-1| + |0|) + (|-1| + |-2|) = 10
- T(3) = 10; T(10) = 524 ;T(100) = 580442; T(10^3) = 583108600.
- Find T(10^6).
- Answer: ?
- Problem 451
- ===========
- Consider the number 15.
- There are eight positive numbers less than 15 which are coprime to 15: 1,
- 2, 4, 7, 8, 11, 13, 14.
- The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11,
- 7, 14
- because
- 1*1 mod 15=1
- 2*8=16 mod 15=1
- 4*4=16 mod 15=1
- 7*13=91 mod 15=1
- 11*11=121 mod 15=1
- 14*14=196 mod 15=1
- Let I(n) be the largest positive number m smaller than n-1 such that the
- modular inverse of m modulo n equals m itself.
- So I(15)=11.
- Also I(100)=51 and I(7)=1.
- Find ∑I(n) for 3≤n≤2·10^7
- Answer: 9848878734a1d751a0e428147ab0b4aa
- Problem 452
- ===========
- Define F(m,n) as the number of n-tuples of positive integers for which the
- product of the elements doesn't exceed m.
- F(10, 10) = 571.
- F(10^6, 10^6) mod 1 234 567 891 = 252903833.
- Find F(10^9, 10^9) mod 1 234 567 891.
- Answer: a75f50818cab61a160cafa2c4145ed23
- Problem 453
- ===========
- A simple quadrilateral is a polygon that has four distinct vertices, has
- no straight angles and does not self-intersect.
- Let Q(m, n) be the number of simple quadrilaterals whose vertices are
- lattice points with coordinates (x,y) satisfying 0 ≤ x ≤ m and 0 ≤ y ≤ n.
- For example, Q(2, 2) = 94 as can be seen below:
- It can also be verified that Q(3, 7) = 39590, Q(12, 3) = 309000 and Q(123,
- 45) = 70542215894646.
- Find Q(12345, 6789) mod 135707531.
- Answer: ?
- Problem 454
- ===========
- In the following equation x, y, and n are positive integers.
- 1 1 1
- ─ + ─ = ─
- x y n
- For a limit L we define F(L) as the number of solutions which satisfy x <
- y ≤ L.
- We can verify that F(15) = 4 and F(1000) = 1069.
- Find F(10^12).
- Answer: cf4e45f50c511e558b3dccb3ed481cb5
- Problem 455
- ===========
- Let f(n) be the largest positive integer x less than 10^9 such that the
- last 9 digits of n^x form the number x (including leading zeros), or zero
- if no such integer exists.
- For example:
- • f(4) = 411728896 (4^411728896 = ...490411728896)
- • f(10) = 0
- • f(157) = 743757 (157^743757 = ...567000743757)
- • Σf(n), 2 ≤ n ≤ 10^3 = 442530011399
- Find Σf(n), 2 ≤ n ≤ 10^6.
- Answer: 22d6cf30a29e14e5c78dca980edc2796
- Problem 456
- ===========
- Define:
- x[n] = (1248^n mod 32323) - 16161
- y[n] = (8421^n mod 30103) - 15051
- P[n] = {(x[1], y[1]), (x[2], y[2]), ..., (x[n], y[n])}
- For example, P[8] = {(-14913, -6630), (-10161, 5625), (5226, 11896),
- (8340, -10778), (15852, -5203), (-15165, 11295), (-1427, -14495), (12407,
- 1060)}.
- Let C(n) be the number of triangles whose vertices are in P[n] which
- contain the origin in the interior.
- Examples:
- C(8) = 20
- C(600) = 8950634
- C(40 000) = 2666610948988
- Find C(2 000 000).
- Answer: e2811a92b4658ca420be740f6c66572b
- Problem 457
- ===========
- Let f(n) = n^2 - 3n - 1.
- Let p be a prime.
- Let R(p) be the smallest positive integer n such that f(n) mod p^2 = 0 if
- such an integer n exists, otherwise R(p) = 0.
- Let SR(L) be ∑R(p) for all primes not exceeding L.
- Find SR(10^7).
- Answer: 5eae79c2f4887f6cf08c099840317a51
- Problem 458
- ===========
- Consider the alphabet A made out of the letters of the word "project":
- A={c,e,j,o,p,r,t}.
- Let T(n) be the number of strings of length n consisting of letters from A
- that do not have a substring that is one of the 5040 permutations of
- "project".
- T(7)=7^7-7!=818503.
- Find T(10^12). Give the last 9 digits of your answer.
- Answer: ?
- Problem 459
- ===========
- The flipping game is a two player game played on a N by N square board.
- Each square contains a disk with one side white and one side black.
- The game starts with all disks showing their white side.
- A turn consists of flipping all disks in a rectangle with the following
- properties:
- • the upper right corner of the rectangle contains a white disk
- • the rectangle width is a perfect square (1, 4, 9, 16, ...)
- • the rectangle height is a triangular number (1, 3, 6, 10, ...)
- Players alternate turns. A player wins by turning the grid all black.
- Let W(N) be the number of winning moves for the first player on a N by N
- board with all disks white, assuming perfect play.
- W(1) = 1, W(2) = 0, W(5) = 8 and W(10^2) = 31395.
- For N=5, the first player's eight winning first moves are:
- Find W(10^6).
- Answer: ?
- Problem 460
- ===========
- On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1)
- for an integer d.
- In each step, the ant at point (x[0], y[0]) chooses one of the lattice
- points (x[1], y[1]) which satisfy x[1] ≥ 0 and y[1] ≥ 1 and goes straight
- to (x[1], y[1]) at a constant velocity v. The value of v depends on y[0]
- and y[1] as follows:
- • If y[0] = y[1], the value of v equals y[0].
- • If y[0] ≠ y[1], the value of v equals (y[1] - y[0]) / (ln(y[1]) -
- ln(y[0])).
- The left image is one of the possible paths for d = 4. First the ant goes
- from A(0, 1) to P[1](1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.
- Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
- From P[1](1, 3) to P[2](3, 3) the ant travels at velocity 3 so the
- required time is 2 / 3 ≈ 0.6667. From P[2](3, 3) to B(4, 1) the ant
- travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required
- time is sqrt(5) / 1.8205 ≈ 1.2283.
- Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.
- The right image is another path. The total required time is calculated as
- 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest
- path for d = 4.
- Let F(d) be the total required time if the ant chooses the quickest path.
- For example, F(4) ≈ 2.960516287.
- We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.
- Find F(10000). Give your answer rounded to nine decimal places.
- Answer: 134fd9e25365ddb970971dd21f386408
- Problem 461
- ===========
- Let f[n](k) = e^k/n - 1, for all non-negative integers k.
- Remarkably, f[200](6) + f[200](75) + f[200](89) + f[200](226) =
- 3.141592644529… ≈ π.
- In fact, it is the best approximation of π of the form
- f[n](a) + f[n](b) + f[n](c) + f[n](d) for n = 200.
- Let g(n) = a^2 + b^2 + c^2 + d^ 2 for a, b, c, d that minimize the error:
- | f[n](a) + f[n](b) + f[n](c) + f[n](d) - π|
- (where |x| denotes the absolute value of x).
- You are given g(200) = 6^2 + 75^2 + 89^2 + 226^2 = 64658.
- Find g(10000). ^
- Answer: 159820276
- Problem 462
- ===========
- A 3-smooth number is an integer which has no prime factor larger than 3.
- For an integer N, we define S(N) as the set of 3-smooth numbers less than
- or equal to N . For example, S(20) = { 1, 2, 3, 4, 6, 8, 9, 12, 16, 18 }.
- We define F(N) as the number of permutations of S(N) in which each element
- comes after all of its proper divisors.
- This is one of the possible permutations for N = 20.
- - 1, 2, 4, 3, 9, 8, 16, 6, 18, 12.
- This is not a valid permutation because 12 comes before its divisor 6.
- - 1, 2, 4, 3, 9, 8, 12, 16, 6, 18.
- We can verify that F(6) = 5, F(8) = 9, F(20) = 450 and F(1000) ≈
- 8.8521816557e21.
- Find F(10^18). Give as your answer its scientific notation rounded to ten
- digits after the decimal point.
- When giving your answer, use a lowercase e to separate mantissa and
- exponent. E.g. if the answer is 112,233,445,566,778,899 then the answer
- format would be 1.1223344557e17.
- Answer: ?
- Problem 463
- ===========
- The function $f$ is defined for all positive integers as follows:
- • $f(1)=1$
- • $f(3)=3$
- • $f(2n)=f(n)$
- • $f(4n + 1)=2f(2n + 1) - f(n)$
- • $f(4n + 3)=3f(2n + 1) - 2f(n)$
- The function $S(n)$ is defined as $\sum_{i=1}^{n}f(i)$.
- $S(8)=22$ and $S(100)=3604$.
- Find $S(3^{37})$. Give the last 9 digits of your answer.
- Answer: 95481696a65b0c1d9f73186a693686f5
- Problem 464
- ===========
- The Möbius function, denoted μ(n), is defined as:
- • μ(n) = (-1)^ω(n) if n is squarefree (where ω(n) is the number of
- distinct prime factors of n)
- • μ(n) = 0 if n is not squarefree.
- Let P(a,b) be the number of integers n in the interval [a,b] such that
- μ(n) = 1.
- Let N(a,b) be the number of integers n in the interval [a,b] such that
- μ(n) = -1.
- For example, P(2,10) = 2 and N(2,10) = 4.
- Let C(n) be the number of integer pairs (a,b) such that:
- • 1 ≤ a ≤ b ≤ n,
- • 99·N(a,b) ≤ 100·P(a,b), and
- • 99·P(a,b) ≤ 100·N(a,b).
- For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319.
- Find C(20 000 000).
- Answer: ?
- Problem 465
- ===========
- The kernel of a polygon is defined by the set of points from which the
- entire polygon's boundary is visible. We define a polar polygon as a
- polygon for which the origin is strictly contained inside its kernel.
- For this problem, a polygon can have collinear consecutive vertices.
- However, a polygon still cannot have self-intersection and cannot have
- zero area.
- For example, only the first of the following is a polar polygon (the
- kernels of the second, third, and fourth do not strictly contain the
- origin, and the fifth does not have a kernel at all):
- Notice that the first polygon has three consecutive collinear vertices.
- Let P(n) be the number of polar polygons such that the vertices (x, y)
- have integer coordinates whose absolute values are not greater than n.
- Note that polygons should be counted as different if they have different
- set of edges, even if they enclose the same area. For example, the polygon
- with vertices [(0,0),(0,3),(1,1),(3,0)] is distinct from the polygon with
- vertices [(0,0),(0,3),(1,1),(3,0),(1,0)].
- For example, P(1) = 131, P(2) = 1648531, P(3) = 1099461296175 and P(343)
- mod 1 000 000 007 = 937293740.
- Find P(7^13) mod 1 000 000 007.
- Answer: ?
- Problem 466
- ===========
- Let P(m,n) be the number of distinct terms in an m×n multiplication table.
- For example, a 3×4 multiplication table looks like this:
- × 1 2 3 4
- 1 1 2 3 4
- 2 2 4 6 8
- 3 3 6 9 12
- There are 8 distinct terms {1,2,3,4,6,8,9,12}, therefore P(3,4) = 8.
- You are given that:
- P(64,64) = 1263,
- P(12,345) = 1998, and
- P(32,10^15) = 13826382602124302.
- Find P(64,10^16).
- Answer: ?
- Problem 467
- ===========
- An integer s is called a superinteger of another integer n if the digits
- of n form a subsequence of the digits of s.
- For example, 2718281828 is a superinteger of 18828, while 314159 is not a
- superinteger of 151.
- Let p(n) be the nth prime number, and let c(n) be the nth composite
- number. For example, p(1) = 2, p(10) = 29, c(1) = 4 and c(10) = 18.
- {p(i) : i ≥ 1} = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...}
- {c(i) : i ≥ 1} = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, ...}
- Let P^D the sequence of the digital roots of {p(i)} (C^D is defined
- similarly for {c(i)}):
- P^D = {2, 3, 5, 7, 2, 4, 8, 1, 5, 2, ...}
- C^D = {4, 6, 8, 9, 1, 3, 5, 6, 7, 9, ...}
- Let P[n] be the integer formed by concatenating the first n elements of
- P^D (C[n] is defined similarly for C^D).
- P[10] = 2357248152
- C[10] = 4689135679
- Let f(n) be the smallest positive integer that is a common superinteger of
- P[n] and C[n].
- For example, f(10) = 2357246891352679, and f(100) mod 1 000 000 007 =
- 771661825.
- Find f(10 000) mod 1 000 000 007.
- Answer: ?
- Problem 468
- ===========
- An integer is called B-smooth if none of its prime factors is greater than
- B.
- Let S[B](n) be the largest B-smooth divisor of n.
- Examples:
- S[1](10) = 1
- S[4](2100) = 12
- S[17](2496144) = 5712
- Define F(n) = ∑[1≤B≤n] ∑[0≤r≤n] S[B](C(n,r)). Here, C(n,r) denotes the
- binomial coefficient.
- Examples:
- F(11) = 3132
- F(1 111) mod 1 000 000 993 = 706036312
- F(111 111) mod 1 000 000 993 = 22156169
- Find F(11 111 111) mod 1 000 000 993.
- Answer: ?
- Problem 469
- ===========
- In a room N chairs are placed around a round table.
- Knights enter the room one by one and choose at random an available empty
- chair.
- To have enough elbow room the knights always leave at least one empty
- chair between each other.
- When there aren't any suitable chairs left, the fraction C of empty chairs
- is determined.
- We also define E(N) as the expected value of C.
- We can verify that E(4) = 1/2 and E(6) = 5/9.
- Find E(10^18). Give your answer rounded to fourteen decimal places in the
- form 0.abcdefghijklmn.
- Answer: 3c2b641262880db5b735cfa4d4c957bc
- Problem 470
- ===========
- Consider a single game of Ramvok:
- Let t represent the maximum number of turns the game lasts. If t = 0, then
- the game ends immediately. Otherwise, on each turn i, the player rolls a
- die. After rolling, if i < t the player can either stop the game and
- receive a prize equal to the value of the current roll, or discard the
- roll and try again next turn. If i = t, then the roll cannot be discarded
- and the prize must be accepted. Before the game begins, t is chosen by the
- player, who must then pay an up-front cost ct for some constant c. For c =
- 0, t can be chosen to be infinite (with an up-front cost of 0). Let R(d,
- c) be the expected profit (i.e. net gain) that the player receives from a
- single game of optimally-played Ramvok, given a fair d-sided die and cost
- constant c. For example, R(4, 0.2) = 2.65. Assume that the player has
- sufficient funds for paying any/all up-front costs.
- Now consider a game of Super Ramvok:
- In Super Ramvok, the game of Ramvok is played repeatedly, but with a
- slight modification. After each game, the die is altered. The alteration
- process is as follows: The die is rolled once, and if the resulting face
- has its pips visible, then that face is altered to be blank instead. If
- the face is already blank, then it is changed back to its original value.
- After the alteration is made, another game of Ramvok can begin (and during
- such a game, at each turn, the die is rolled until a face with a value on
- it appears). The player knows which faces are blank and which are not at
- all times. The game of Super Ramvok ends once all faces of the die are
- blank.
- Let S(d, c) be the expected profit that the player receives from an
- optimally-played game of Super Ramvok, given a fair d-sided die to start
- (with all sides visible), and cost constant c. For example, S(6, 1) =
- 208.3.
- Let F(n) = ∑[4≤d≤n] ∑[0≤c≤n] S(d, c).
- Calculate F(20), rounded to the nearest integer.
- Answer: ?
- Problem 471
- ===========
- The triangle ΔABC is inscribed in an ellipse with equation $\frac {x^2}
- {a^2} + \frac {y^2} {b^2} = 1$, 0 < 2b < a, a and b integers.
- Let r(a,b) be the radius of the incircle of ΔABC when the incircle has
- center (2b, 0) and A has coordinates $\left( \frac a 2, \frac {\sqrt 3} 2
- b\right)$.
- For example, r(3,1) = ½, r(6,2) = 1, r(12,3) = 2.
- Let $G(n) = \sum_{a=3}^n \sum_{b=1}^{\lfloor \frac {a - 1} 2 \rfloor} r(a,
- b)$
- You are given G(10) = 20.59722222, G(100) = 19223.60980 (rounded to 10
- significant digits).
- Find G(10^11).
- Give your answer in scientific notation rounded to 10 significant digits.
- Use a lowercase e to separate mantissa and exponent.
- For G(10) the answer would have been 2.059722222e1.
- Answer: ?
- Problem 472
- ===========
- There are N seats in a row. N people come one after another to fill the
- seats according to the following rules:
- 1. No person sits beside another.
- 2. The first person chooses any seat.
- 3. Each subsequent person chooses the seat furthest from anyone else
- already seated, as long as it does not violate rule 1. If there is
- more than one choice satisfying this condition, then the person
- chooses the leftmost choice.
- Note that due to rule 1, some seats will surely be left unoccupied, and
- the maximum number of people that can be seated is less than N (for N >
- 1).
- Here are the possible seating arrangements for N = 15:
- We see that if the first person chooses correctly, the 15 seats can seat
- up to 7 people.
- We can also see that the first person has 9 choices to maximize the number
- of people that may be seated.
- Let f(N) be the number of choices the first person has to maximize the
- number of occupants for N seats in a row. Thus, f(1) = 1, f(15) = 9,
- f(20) = 6, and f(500) = 16.
- Also, ∑f(N) = 83 for 1 ≤ N ≤ 20 and ∑f(N) = 13343 for 1 ≤ N ≤ 500.
- Find ∑f(N) for 1 ≤ N ≤ 10^12. Give the last 8 digits of your answer.
- Answer: ?
- Problem 473
- ===========
- Let $\varphi$ be the golden ratio: $\varphi=\frac{1+\sqrt{5}}{2}.$
- Remarkably it is possible to write every positive integer as a sum of
- powers of $\varphi$ even if we require that every power of $\varphi$ is
- used at most once in this sum.
- Even then this representation is not unique.
- We can make it unique by requiring that no powers with consecutive
- exponents are used and that the representation is finite.
- E.g: $2=\varphi+\varphi^{-2}$ and $3=\varphi^{2}+\varphi^{-2}$
- To represent this sum of powers of $\varphi$ we use a string of 0's and
- 1's with a point to indicate where the negative exponents start.
- We call this the representation in the phigital numberbase.
- So $1=1_{\varphi}$, $2=10.01_{\varphi}$, $3=100.01_{\varphi}$ and
- $14=100100.001001_{\varphi}$.
- The strings representing 1, 2 and 14 in the phigital number base are
- palindromic, while the string representating 3 is not.
- (the phigital point is not the middle character).
- The sum of the positive integers not exceeding 1000 whose phigital
- representation is palindromic is 4345.
- Find the sum of the positive integers not exceeding $10^{10}$ whose
- phigital representation is palindromic.
- Answer: a4ea7a2040b6385b6d12863fd693e434
- Problem 474
- ===========
- For a positive integer n and digits d, we define F(n, d) as the number of
- the divisors of n whose last digits equal d.
- For example, F(84, 4) = 3. Among the divisors of 84 (1, 2, 3, 4, 6, 7, 12,
- 14, 21, 28, 42, 84), three of them (4, 14, 84) have the last digit 4.
- We can also verify that F(12!, 12) = 11 and F(50!, 123) = 17888.
- Find F(10^6!, 65432) modulo (10^16 + 61).
- Answer: ?
- Problem 475
- ===========
- 12n musicians participate at a music festival. On the first day, they form
- 3n quartets and practice all day.
- It is a disaster. At the end of the day, all musicians decide they will
- never again agree to play with any member of their quartet.
- On the second day, they form 4n trios, each musician avoiding his previous
- quartet partners.
- Let f(12n) be the number of ways to organize the trios amongst the 12n
- musicians.
- You are given f(12) = 576 and f(24) mod 1 000 000 007 = 509089824.
- Find f(600) mod 1 000 000 007.
- Answer: 75780067
- Problem 476
- ===========
- Let R(a, b, c) be the maximum area covered by three non-overlapping
- circles inside a triangle with edge lengths a, b and c.
- Let S(n) be the average value of R(a, b, c) over all integer triplets (a,
- b, c) such that 1 ≤ a ≤ b ≤ c < a + b ≤ n
- You are given S(2) = R(1, 1, 1) ≈ 0.31998, S(5) ≈ 1.25899.
- Find S(1803) rounded to 5 decimal places behind the decimal point.
- Answer: 4d6a99b2a0f22af561aeeb69c0126fef
Add Comment
Please, Sign In to add comment
