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Homework: Math for Developers

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Aug 17th, 2015
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  1.  
  2. Problem 1. Some Primes
  3. Find the 24th, 101st and 251st prime number.
  4.  
  5. Solution:
  6. 24th prime number is 89
  7. 101st prime number is 547
  8. 251st prime number is 1597
  9. - Source for solution: http://www.wolframalpha.com/
  10.  
  11. Problem 2. Some Fibonacci Primes
  12. Check if the 24th, 101st and 251st prime numbers are part of the base Fibonacci number set. What is their position?
  13. Solution:
  14. 24th prime number ( 89 ) is part of the base Fibonacci number set. Its position is: 11th
  15. 101st prime number is not part of the base Fibonacci number set;
  16. 251th prime number ( 1597 ) is part of the base Fibonacci number set. Its position is: 17th
  17. - Source for solution: http://www.wolframalpha.com/
  18.  
  19. Problem 3. Some Factorials
  20. Find 100!, 171! and 250! Give all digits.
  21. Solution:
  22. 100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
  23.  
  24. 171! = 1241018070217667823424840524103103992616605577501693185388951803611996075221691752992751978120487585576464959501670387052809889858690710767331242032218484364310473577889968548278290754541561964852153468318044293239598173696899657235903947616152278558180061176365108428800000000000000000000000000000000000000000
  25.  
  26. 250! =
  27. 3232856260909107732320814552024368470994843717673780666747942427112823747555111209488817915371028199450928507353189432926730931712808990822791030279071281921676527240189264733218041186261006832925365133678939089569935713530175040513178760077247933065402339006164825552248819436572586057399222641254832982204849137721776650641276858807153128978777672951913990844377478702589172973255150283241787320658188482062478582659808848825548800000000000000000000000000000000000000000000000000000000000000
  28. - Source for solution: http://www.wolframalpha.com/
  29.  
  30. Problem 4. Calculate Hypotenuse
  31. You are given three right angled triangles. Find the length of their hypotenuses.
  32. 1. Catheti: 3 and 4
  33. Solution:
  34. Hypotenuses = √(32+42)= 5
  35. 2. Catheti: 10 and 12
  36. Solution:
  37. Hypotenuses = √(102+122)= 15.6204993518
  38. 3. Catheti 100 and 250
  39. Solution:
  40. Hypotenuses = √(1002+2502)= 269.258240357
  41.  
  42. Problem 5. Numeral System Conversions
  43. Convert 1234d to binary and hexadecimal numeral systems.
  44. Solution:
  45. 1234d = 10011010010b=4D2hex
  46.  
  47. Convert 1100101b to decimal and hexadecimal numeral systems.
  48. Solution:
  49. 1100101b=101d=65hex
  50.  
  51. Convert ABChex to decimal and binary numeral systems.
  52. Solution:
  53. ABChex=2748d=101010111100b
  54.  
  55. - Source for solution: Windows 7 - Calculator
  56.  
  57. Problem 6. Least Common Multiple
  58. Find LCM(1234, 3456).
  59. Solution:
  60. LCM(1234, 3456) = 2 132 352
  61.  
  62. - Source for solution: http://www.wolframalpha.com/
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