If p and r are integers, and p^2=28r, then r must be...

swerve wrote:If p and r are integers, and p^2=28r, then r must be divisible by which of the following?

A. 2
B. 4
C. 5
D. 7
E. 14

The OA is D.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

$$28 = 2^2 * 7$$

So we can rewrite our initial equation as $$p^2 = 2^2 *7r$$
The only way the above can be expressed as a perfect square is if r = 7, (or 7 raised to some odd exponent. You can think of a perfect square as an integer whose prime factorization will yield a scenario in which every prime base is raised to a multiple of 2.)
Put another way, if r = 7, then we'd have $$p^2 = 2^2 *7^2 = (2*7)^2$$

So we know that r must be divisible by 7. The answer is D