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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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19 Aug 2015, 23:19
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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
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19 Aug 2015, 23:41
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
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19 Aug 2015, 23:58
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
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20 Aug 2015, 07:58
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
Answer: D) 7



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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi
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20 Aug 2015, 08:14
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
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21 Aug 2015, 06:32
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. Answer D



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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi
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23 Aug 2015, 11:29
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
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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. Answer: D
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Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi
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