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

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by VJesus12 » Sun Apr 14, 2019 9:58 am
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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
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by Scott@TargetTestPrep » Fri Apr 26, 2019 2:36 pm
VJesus12 wrote: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

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