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25 May 2021, 02:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
67% (00:57) correct
33%
(01:42)
wrong
based on 123
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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
(A) 2
(B) 4
(C) 5
(D) 7
(E) 14
25 May 2021, 02:52
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D) 7
28= 2^2*7 . Since p is a square of a number then r must be 7k^2. So 7 must be factor of r.
Posted from my mobile device
28= 2^2*7 . Since p is a square of a number then r must be 7k^2. So 7 must be factor of r.
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29 May 2021, 06:00
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Bunuel
OFFICAL EXPLANATION
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.
