Ian Stewart wrote:

kanha81 wrote:If p and q are prime numbers, how many divisors does the product
(p^3) * (q^6) have?

(A) 9
(B) 12
(C) 18
(D) 28
(E) 363

If p and q are *different* primes, then the answer will be 28. We don't need to list of all of the divisors here - that's a bit time consuming - but if you're interested, they are:

1, q, q^2, q^3, q^4, q^5, q^6
p, pq, pq^2, pq^3, pq^4, pq^5, pq^6
p^2, (p^2)q, (p^2)q^2, (p^2)q^3, (p^2)q^4, (p^2)q^5, (p^2)q^6
p^3, (p^3)q, (p^3)q^2, (p^3)q^3, (p^3)q^4, (p^3)q^5, (p^3)q^6

That's just all of the combinations of products (p^a)(q^b), where a = 0, 1, 2 or 3, and b = 0, 1, 2, 3, 4, 5, or 6. Because we have 4 choices for a, and 7 choices for b, we have 4x7 = 28 divisors in total. Similarly, you can count all the divisors of a number very quickly once you have its prime factorization:

-look only at the exponents
-add one to each exponent, and multiply the resulting numbers together

Here we have (p^6)(q^3). Add one to each power and multiply: 7x4 = 28.

___________

It is genuinely important that the primes p and q are different, something the question above does not make clear. If p and q are not different -- that is, if p = q -- then (p^6)(q^3) = p^9, which has only ten divisors. The question is poorly worded, at least if that's the wording from the original source. Where is it from?