$x=9^{10}-3^{17}$ 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?

I got three possible values for n. I factored x as 3^{16}(3^2 - 3) = 3^{16} * 6 = 2 * 3^{17}. The primes are 2 and 3, so the only n with exactly two factors are 2 and 3. But wait, the question asks for n such that x/n is an integer, so n must be a factor. That gives only two. Where does the third come from?

Tricky question. I initially thought n could be any prime factor, but then I realized n must have exactly two factors, which means n is prime. So we just need the number of distinct prime factors of x. Factoring x as 3^{17}(3^3 - 1) = 3^{17} * 26 gives primes 2, 13, and 3. So three primes, hence three possible n. Answer C.

How do we know that n must be prime? I understand that a number with exactly two factors is prime, but is it possible that n could be something like p^2? No, p^2 has three factors. So n must be prime. That's the key. So we just need the number of distinct prime factors. Good.

I got this wrong because I didn't simplify 9^{10} to 3^{20}. I just left it as 9^{10} - 3^{17} and couldn't factor. Lesson: always convert to common bases when possible. Once I did that, it became 3^{17}(27-1)=3^{17}*26, which is straightforward. Difficulty is around 700 I'd say.