Verified answer

Since τ is multiplicative, think of the prime factorizations of n.

Since τ(p^k) = k+1, it suffices to look at factorizations of 15:

15 (no factorization) ==> k+1 = 15 ==> k = 14.

So, τ(p^14) = 15 for any prime p.

3 * 5 ==> j+1 = 3 and k+1 = 5 ==> j = 2 and k = 4.

So, τ(p^2 q^4) = 3 * 5 = 15 for any distinct primes p, q.

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I hope this helps!

Definitely follow kb's train of logic -- that will get you the answer.

But also, don't neglect the other way to form 15 from factors -- the "null factorization" -- that is, just as you can write

15 = 3•5,

which leads to numbers, n that are of the form,

n = p^2 • q^4,

you can also write

15 = 15,

which leads to numbers, n that are of the form,

n = p^14