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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If p and r are integers, and p^2=28r, then r must be... tagged by: swerve ##### This topic has 2 expert replies and 0 member replies ### Top Member ## If p and r are integers, and p^2=28r, then r must be... 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. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 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 _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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### GMAT/MBA Expert

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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, letâ€™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.

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