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# If p and r are integers, and p^2 =28r, then r must be divisible by whi

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If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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19 Aug 2015, 23:19
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77% (01:06) correct 23% (01:35) wrong based on 137 sessions

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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

Kudos for a correct solution.

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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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19 Aug 2015, 23:41
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p^2 = 28r
p= +- 2 * $$\sqrt{(7r)}$$

Of the values given in the options, p will be an integer only if r = 7

Ans:D
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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19 Aug 2015, 23:58
1
Bunuel 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

Ans: D

Solution: for p to be an int 28 r must be whole square of a number.
28r= 7*2*2*r
to make it whole square we need 7
so r can must be divisible by 7y where y is itself a whole square.
so D is the ans
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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20 Aug 2015, 07:58
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If p and r are integers, and p^2 = 28r, then r must be divisible by which of the following?

28 has 2 twos and 1 seven. That means r must contain a 7 for 28r to be divisible by p

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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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20 Aug 2015, 08:14
1
Bunuel 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

Kudos for a correct solution.

28r is a square number
2^2*7r is a square number
Therefore, r has to be 7.
7 is divisible by 1 as well as 7
Therefore, the correct option is D
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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21 Aug 2015, 06:32
1
Bunuel 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

Kudos for a correct solution.

P^2=28r and 28=2^2*7
Therefore, since p^2 is a square, r needs to have a 7.
Math Expert
Joined: 02 Sep 2009
Posts: 59721
Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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23 Aug 2015, 11:29
1
Bunuel 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

Kudos for a correct solution.

Economist GMAT Tutor Official Solution:

First, we want to know for what, precisely, the question asks. This is a “must be” question. We can see that there are lots of possible values for p and r, but we are asked to find the factor (from among the answer choices) that must be present regardless of our choice of p or r.

So, what information is available to help us solve this problem?

(1) p and r are integers
(2) p^2=28r

From (1), we know that p^2 is a perfect square, and by extension, we know that 28r is a perfect square.

What do we know about perfect squares and their prime factors? Each must have a partner!

So, if we factor 28r, we get 2 • 2 • 7 • r. So, whatever other factors r contains, it must, at least, have the partner for 7. So, r must be divisible by 7.
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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05 Jun 2017, 16:15
Bunuel 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.

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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi  [#permalink]

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20 Mar 2019, 22:16
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi   [#permalink] 20 Mar 2019, 22:16
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