Hi VJesus12,

We're told that A = 5^15 - 625^3 and A/X is an integer. X is a positive integer greater than 1, such that it does NOT have a factor P such that 1 < P < X, we're asked how many different values for X are possible. While this question is 'quirky looking', it's based on Exponent rules and Prime Factorization, so we'll have to 'rewrite' some of the information that we've been given. Since X has NO factors other than itself and 1, then that means that X must be PRIME.

To start, 625 = (5^3), so we can rewrite 625^3 as (5^3)^3 = 5^9

Thus, A = 5^15 - 5^9. This can be factored down into....
5^15 - 5^9 =
(5^9)(5^6 - 1) =
(5^9)(5^3 - 1)(5^3 + 1) =
(5^9)(124)(126) =
(5^9)(4)(31)(2)(63) =
(5^9)(2)(2)(31)(2)(7)(3)(3)

Since X must be PRIME and a factor of A, then X could be 2, 3, 7 or 31 --> 4 possible values

Final Answer: E

GMAT assassins aren't born, they're made,
Rich