x=9^10–3^17 and x/n is an integer. If n is a positive

[VJesus12]

x=9^10-3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

[Scott@TargetTestPrep]

x=9^10-3^17 and x/n is an integer. If n is a positive integer that has exactly two factors, how many different values for n are possible?

(A) One
(B) Two
(C) Three
(D) Four
(E) Five

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT

Since only prime numbers have exactly two factors, we see that n is a prime, and therefore, we can rephrase the question as "how many prime factors does x have?" So we need to prime factorize x:

x = 9^10 - 3^17 = (3^2)^10 - 3^17 = 3^20 - 3^17 = 3^17(3^3 - 1) = 3^17(26) = 3^17 * 2 * 13

We see that x has three prime factors.

Answer: C