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

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by Sionainn@PrincetonReview » Mon May 21, 2018 10:34 am
This is a good one for breaking numbers down into their basic building blocks by prime factoring. In this case 28 = 2*2*7. If p and r are both integers when you factor the value on the right side, it should break down so that there are pairs of each number (to match the squaring on the right side). With 28, we already have a pair of 2s but 7 needs a mate to make it a pair. So r must bring a 7 to the party, so it must be divisible by 7 and the answer is D.

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by Vincen » Tue May 22, 2018 12:13 am
VJesus12 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 .

How can I find the correct answer here? Is there someone who can help me? Thanks in advance.
Hello Vjesus12.

We have $$p^2=28r\ \ \ \Rightarrow\ \ p=\pm\ \sqrt{28r}=\sqrt{4\cdot7r}=2\sqrt{7r}.$$ Since p is an integer, then r must have an odd number of 7's in its prime factorization.

Therefore, r must be divisible by 7. Hence, the correct answer is the option D.

I hope it helps. <i class="em em-smiley"></i>
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by Scott@TargetTestPrep » Thu May 24, 2018 12:33 pm
VJesus12 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 see that 28r must be a perfect square, which means that 28r must break down to unique prime factors that are in even quantities.

Since 28 = 2^2 x 7, r must contain at least one factor of 7.

Answer: D

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