Ian Stewart wrote:

smclean23 wrote:If n and p are different positive prime numbers, which of the integers n^4, p^3 , and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np

If n and p are different primes:

The divisors of p^3 will be 1, p, p^2 and p^3 (four divisors).

The divisors of n^4 will be 1, n, n^2, n^3 and n^4 (five divisors).

The divisors of np will be 1, n, p and np (four divisors).

You don't need to do all of that if you know how to count divisors from a prime factorization. If you have a prime factorization of x, then you can count the number of divisors of x:

-look only at the powers in the prime factorization
-add one to each power and multiply.

So the number p^3 has 3+1 = 4 divisors, and np has (1+1)(1+1) = 2*2 = 4 divisors. A few more examples- if p and q are different primes, the number (p)*(q^5) would have (1+1)*(5+1) = 2*6 = 12 divisors. The number (p^99)*(q^99) would have (99+1)*(99+1) = 100*100 = 10,000 divisors.