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# a = 5^15 - 625^3 and a/x is an integer, where x is a . . . .

tagged by: VJesus12

### Top Member

#### a = 5^15 - 625^3 and a/x is an integer, where x is a . . . .

a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer greater than 1, such that it does NOT have a factor p such that 1 < p < x, then how many different values for x are possible?

A. None
B. One
C. Two
D. Three
E. Four

The OA is the option D.

I am not sure about how do I have to solve this PS question? Experts, may you help me here? Thanks in advanced.

### GMAT/MBA Expert

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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

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

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Contact Rich at Rich.C@empowergmat.com

### GMAT/MBA Expert

GMAT Instructor
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Quote:
a = 5^15 - 625^3 and a/x is an integer, where x is a positive integer greater than 1, such that it does NOT have a factor p such that 1 < p < x, then how many different values for x are possible?

A. None
B. One
C. Two
D. Three
E. Four
We can start by simplifying a:

5^15 - 625^3 = 5^15 - (5^4)^3 = 5^15 - 5^12 = 5^12(5^3 - 1) = 5^12(124) = 5^12(4)(31) = 5^12( 2^2)(31)

If a/x is an integer, then x is a factor of a. However, if x does not have a factor p such that 1 < p < x, then x must be prime number. For example, if x = 5, we see that x doesn’t have a factor between 1 and itself. Since a has three distinct prime factors, there are three distinct values for x: 2, 5 and 31.

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