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

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

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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.

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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

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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
We are given that p^2 = 28r, which means that 28r is a perfect square. We must remember that all perfect squares break down to unique prime factors, each of which has an exponent that is a multiple of 2. So, let’s break down 28 into its prime factors to determine the minimum value of r.

28 = 7 x 4 = 7 x 2^2
In order to make 28r a perfect square, the smallest value of r is 7^1, so that 28r = x 3^1) = 2^2 x 7^2, which is a perfect square.

Answer: D

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scott@targettestprep.com



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