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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
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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-------------------------------------------------------------------- The Mind is Everything, What we Think we Become.
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.
Re: If p and r are integers, and p^2 =28r, then r must be divisible by whi
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20 Mar 2019, 22:16
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