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Bellazon

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Posted

This one is tougher... I believe each integer can be represented in a way so that

i = p1^e1 * p2^e2* p3^e3... * pn^en

where i is the given integer, p is a set of unique prime numbers, and e is the exponent.

For example:

1 = 2^0

2 = 2^1

3 = 3^1

4 = 2^2

5 = 5^1

6 = 2^1 * 3^1

7 = 7^1

8 = 2^3

9 = 3 ^ 2

...

96 = 2^5 * 3^1

...

160 = 2^5 * 5^1

...

115200 = 2^9 * 3^2 * 5^2

Ready?

Set! Go!

Posted

10 = 2^1 * 5^1

11 = 11^1

12 = 2^2 * 3^1

13 = 13^1

14 = 2^1 * 7^1

15 = 3^1 * 5^1

16 = 2^4

17 = 17^1

18 = 2^1 * 3^2

19 = 19^1

20 = 2^2 * 5^1

21 = 3^1 * 7^1

22 = 2^1 * 11^1

23 = 23^1

24 = 2^3 * 3^1

25 = 5^2

26 = 2^1 * 13^1

27 = 3^3

28 = 2^2 * 7^1

29 = 29^1

30 = 2^1 * 3^1 * 5^1

Posted

41 = PRIME!

42 = 2^1 * 3^1 * 7^1

43 = PRIME!

44 = 2^2 * 11^1

45 = 3^2 * 5^1

46 = 2^1 * 23^1

47 = PRIME!

48 = 2^4 * 3^1

49 = 7^2

50 = 2^1 * 5^2

Posted

Keep up with me folks! :D

51 = 3^1 * 17^1

52 = 2^2 * 13^1

53 = PRIME!

54 = 2^1 * 3^3

55 = 5^1 * 11^1

56 = 2^3 * 7^1

57 = 3^1 * 19^1

58 = 2^1 * 29^1

59 = PRIME!

60 = 2^2 * 3^1 * 5^1

Posted
31 = 31^1

32 = 2^4

33 = 11^1* 3^1

34 = 17^1*2^1

:wacko: This is too complicated for me

No no no... this is just a nerdy way to do factor trees!

60 = 2 * 2 * 5 *3

since 2 * 2 = 2 ^2 and n^1 = n

we can write it like this

60 = 2^2 * 3^1 * 5^1

also... since n^0 = 1

we can throw in more useless numbers to make ourselves look smarter

60 = 2^2 * 3^1 * 5^1 * 7^0 * 11^0 * 13^0

Posted

71 = PRIME

72 = 2^1 * 3^1 * 2^1 * 3^1 *2^1

73 = PRIME

74 = 37^1 * 2^1

75 = 5^1 * 3^1 * 5^1

76 = 2^1 * 19^1 * 2^1

77 = 11^1 * 7^1

78 = 39^1 * 2^1

79 = PRIME

80 = 2^1 * 2^2 * 2^1 * 5^1

  • 2 weeks later...

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