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If a, m and n are positive integers, is n^(2a) a multiple of

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If a, m and n are positive integers, is n^(2a) a multiple of [#permalink] New post 30 Apr 2013, 04:33
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If a, m and n are positive integers, is n^(2a) a multiple of m^a?

(1) n is a multiple of m/2
(2) n is a multiple of 2m
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Apr 2013, 04:40, edited 1 time in total.
Edited the question.
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Re: If a, m and n are positive integers, is n^(2a) a multiple of [#permalink] New post 30 Apr 2013, 05:00
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If a, m and n are positive integers, is n^(2a) a multiple of m^a?

The question asks whether \frac{n^{2a}}{m^a}=(\frac{n^2}{m})^a=integer. So, the question basically asks whether n^2 is a multiple of m.

(1) n is a multiple of m/2. If n=3 and m=2, then n^2=9 is NOT a multiple of m=2 but if n=2 and m=2, then n^2=4 IS a multiple of m=2. Not sufficient.

(2) n is a multiple of 2m. This implies that n is a multiple of m, thus n^2 is also a multiple of m. Sufficient.

Answer: B.

Hope it's clear.
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Re: If a, m and n are positive integers, is n^(2a) a multiple of [#permalink] New post 30 Apr 2013, 05:21
Option B.

From Stmt 1:
n is a multiple of m/2 . case 1:n= 9 , m=6 and a =1
9 is a multiple of 3(m/2) bur 81 is not a multiple of 6.

case 2 : n=8 , m=4 and a=1.
8 is a multiple of 2(m/2) and 8^2 = 64 is a multiple of 4.
so insufficient.

stmt 2:

n=2m*k. now both sides raised to the power or 2a gives
n^2a = 2^2a * k^2a * m^2a => 2^2a* k^2a * m^a* m^a

Hence it is a multiple of m^a. Sufficient
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Re: If a, m and n are positive integers, is n^(2a) a multiple of [#permalink] New post 06 Jul 2014, 21:27
The question basically asks if n^2a/m^a =integer,is (n^a.n^a)/m^a =integer,is (n/m)^a.n^a=integer,since n and a are positive integer,it will always be an integer,so the question really is n a multiple of m then only (n/m)^a.n^a=integer

Statement 1
n/(m/2)=integer↪2n/m=integer↪n=1,m=2 then n/m≠integer↪↪but n=2,m=1,then n/m=integer…not sufficient

Statement 2

n/2m=integer↪n/m=2*integer….sufficient

Answer-B
Re: If a, m and n are positive integers, is n^(2a) a multiple of   [#permalink] 06 Jul 2014, 21:27
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