If n is multiple of 5, and n = p²q, where p and q are prime numbers, which of the following MUST be a multiple of 25?

A) p²
B) q²
C) pq
D) p²q²
E) p³q

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k (i.e., N is a multiple of k), then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

If p and q are prime numbers, and p²q is divisible by 5, then either p = 5, q = 5 or they both equal 5.

We're looking for an expression that MUST be divisible by 25, which means there must be TWO 5's "hiding" in the prime factorization of the expression.

A) p²
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 2 and q = 5, in which case p² is NOT divisible by 25
ELIMINATE A

B) q²
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 5 and q = 2, in which case q² is NOT divisible by 25
ELIMINATE B

C) pq
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 5 and q = 2, in which case pq is NOT divisible by 25
ELIMINATE C

D) p²q²
YES, we can be certain that there are TWO 5's "hiding" in this expression.
If p = 5, then p²q² = 25q², which is DEFINITELY divisible by 25
If q = 5, then p²q² = 25p², which is DEFINITELY divisible by 25

E) p³q
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 2 and q = 5, in which case p³q is NOT divisible by 25
ELIMINATE E

Answer = D

Cheers,
Brent