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Updated on: 29 Mar 2020, 13:23
Originally posted by cristinaolt19 on 23 Jan 2020, 06:33.
Last edited by Bunuel on 29 Mar 2020, 13:23, edited 1 time in total.
Last edited by Bunuel on 29 Mar 2020, 13:23, edited 1 time in total.
Renamed the topic and edited the question.
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Difficulty:
35%
(medium)
Question Stats:
67% (01:29) correct
33%
(01:31)
wrong
based on 172
sessions
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Date
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If x is a positive integer such that x^2 is a multiple of 56, what is the least possible value of x?
A. 14
B. 22
C. 26
D. 28
E. 56
A. 14
B. 22
C. 26
D. 28
E. 56
23 Jan 2020, 06:42
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cristinaolt19
x^2 = 56k
x^2 = 2^3*7*k
k must make powers of 2 and 7 even, so that 2^3*7*k becomes a perfect square. The least value of k is therefore 2*7.
So, x^2 = 2^3*7*k = 2^4*7^2.
x = 2^2*7.
P.S. We have specific rules of posting a question on the forum: https://gmatclub.com/forum/rules-to-pos ... 08733.html Please read and edit the first post: provide the source, options and the OA. Thank you.
globaldesi
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28 Jan 2020, 20:28
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cristinaolt19
\(56 = 8*7\)
or \(2^3 *7\)
thus x^2 should atleast have factors of the form \(2^3 *7\)
and since x an integer \(x^2\) should be a perfect square
thus \(x^2\) should be of the form \(2^4 *7^2\)
or\( x = 2^2*7\)
= 28
