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Oodles of Factors II (Posted on 2010-10-11) Difficulty: 3 of 5
A. What is the lowest base 12 positive integer that has exactly 10 (base 12) distinct positive factors?

B. Exactly 1,000 (base 12) distinct positive factors?

C. Exactly 1,000,000 (base 12) distinct positive factors?

For example, the distinct positive factors of 40 (base 12) are the base 12 numbers 1, 2, 3, 4, 6, 8, 10, 14, 20, and 40. Accordingly, 40 (base 12) has precisely A (base 12) distinct positive factors.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts thoughts on part B | Comment 2 of 7 |

1000 (base 12) = 1728 (base 10), i.e., 12^3 (base 10)

1728 = 2*2*2*2*2*2*3*3*3

so at one extreme we could have the first power of six different primes multiplied by the squares of three other primes, or on the other hand, one prime (presumably 2) raised to the 1727 power.

By taking the squares of 2, 3 and 5 and the first powers of 7, 11, 13, 17, 19 and 23, we get 6,692,786,100 (all decimal). The other extreme of 2^1727 is a 520-digit number in base 10.

However, if we go with three cubes and three squares we get 27,054,027,000.

How about two 7th powers and three squares?: 41,493,513,600.

How about an 8th power, a square and six first powers?: 3,724,680,960, the smallest so far. I don't see a way of improving this, but there might be.

In base 12, the decimal 2^8 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19 = 3,724,680,960 is 87B,47A,800. It's the lowest I've found.


  Posted by Charlie on 2010-10-11 13:35:45
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