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18 Jul 2018, 02:46
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
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65% (01:23) correct
35%
(01:43)
wrong
based on 316
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GMAT CLUB TESTS' FRESH QUESTION
If both m and n are positive integers, is m a multiple of n?
(1) m + 3n is a multiple of n
(2) 2m + n is a multiple of n
M36-54
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08 Nov 2021, 06:13
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Bunuel
Official Solution:
If both \(m\) and \(n\) are positive integers, is \(m\) a multiple of \(n\) ?
(1) \(m + 3n\) is a multiple of \(n\)
\(3n\) is obviously a multiple of \(n\), thus for the sum of \(m\) and \(3n\) to be a multiple of \(n\), \(m\) must also be a multiple of \(n\). Sufficient.
(2) \(2m + n\) is a multiple of \(n\)
\(n\) is obviously a multiple of \(n\), thus for the sum of \(2m\) and \(n\) to be a multiple of \(n\), \(2m\) must also be a multiple of \(n\). Does \(2m\) being a multiple of \(n\) imply that \(m\) is a multiple of \(n\) ? Not necessarily! For example, if \(m=3\) and \(n=2\), then the answer is NO but if \(m=2\) and \(n=2\), then the answer is YES. Not sufficient.
Answer: A
General Discussion
18 Jul 2018, 08:41
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To find whether m is a multiple of n ?
Statement 1
m + 3n is a multiple of n
=> (m + 3n) = n * C (C is some integer)
=> m = nC - 3n
=> m = n(C - 3n)
m is a multiple of n
Statement 1 is sufficient
Statement 2
2m + n is a multiple of n
=> 2m + n = n * C (C is some integer)
=> 2m = n * C - n
=> 2m = n(C - 1)
=> m = \(\frac{n(C-1)}{2}\)
We don't know the divisibility of either "n" or "C-1" with respect to "2"
Hence we can't say m is a multiple of n
Statement 2 is insufficient
Hence option A
Statement 1
m + 3n is a multiple of n
=> (m + 3n) = n * C (C is some integer)
=> m = nC - 3n
=> m = n(C - 3n)
m is a multiple of n
Statement 1 is sufficient
Statement 2
2m + n is a multiple of n
=> 2m + n = n * C (C is some integer)
=> 2m = n * C - n
=> 2m = n(C - 1)
=> m = \(\frac{n(C-1)}{2}\)
We don't know the divisibility of either "n" or "C-1" with respect to "2"
Hence we can't say m is a multiple of n
Statement 2 is insufficient
Hence option A
