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# If k is a multiple of 3 and k = (m^2)n, where m and n are

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If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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04 Feb 2013, 13:34
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65% (02:06) correct 35% (01:07) wrong based on 267 sessions

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If k is a multiple of 3 and $$k = (m^2)n$$, where m and n are prime numbers, which of the following must be a multiple of 9?

(A) $$m^2$$
(B) $$n^2$$
(C) $$mn$$
(D) $$mn^2$$
(E) $$(mn)^2$$

Source: Gmat Hacks 1800
[Reveal] Spoiler: OA

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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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04 Feb 2013, 21:19
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Expert's post
megafan wrote:
If k is a multiple of 3 and $$k = (m^2)n$$, where m and n are prime numbers, which of the following must be a multiple of 9?

(A) $$m^2$$
(B) $$n^2$$
(C) $$mn$$
(D) $$mn^2$$
(E) $$(mn)^2$$

Source: Gmat Hacks 1800

Given prime factorization of k:
$$k = (m^2)n$$,
If k is a multiple of 3, we can say that either m = 3 or n = 3.
So either m^2 or n^2 will be a multiple of 9 but we don't know which of them is a multiple of 9.
That is why none of A, B, C and D work.
(E) $$(mn)^2$$ includes both$$m^2$$ and $$n^2$$ and hence it must be a multiple of 9.
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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04 Feb 2013, 14:05
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Since 'm' and 'n' are primes, In order for k to be a multiple of 3 either 'n' or 'm' must be 3. Also, In order to be a multiple of 9 the product must have two 3's as factors.
A) No. 'm' could be 3 or 'n' could be 3 we don't know which one
B) No. Same logic as above.
C) No. 'm' and 'n' could both be 3 in which case it would work. But they could also be '5' and '7' or some other pair of primes.
D) No. n^2 could be 3^2 or it could be 5^2 (or some other prime number)
E) Correct Answer. Since 3 must be one of the numbers, squaring the product must yield 9 as a factor.
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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05 Feb 2013, 03:01
1
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Expert's post
megafan wrote:
If k is a multiple of 3 and $$k = (m^2)n$$, where m and n are prime numbers, which of the following must be a multiple of 9?

(A) $$m^2$$
(B) $$n^2$$
(C) $$mn$$
(D) $$mn^2$$
(E) $$(mn)^2$$

Source: Gmat Hacks 1800

This question is almost exact copy of the following GMAT Prep question: if-is-n-is-multiple-of-5-and-n-p-2-q-where-p-and-q-are-prim-92383.html
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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04 Feb 2013, 14:28
megafan wrote:
If k is a multiple of 3 and $$k = (m^2)n$$, where m and n are prime numbers, which of the following must be a multiple of 9?

(A) $$m^2$$
(B) $$n^2$$
(C) $$mn$$
(D) $$mn^2$$
(E) $$(mn)^2$$

Source: Gmat Hacks 1800

must be a multiple of 9
m and n are prime...lets have 2 and 3
now either of then can be a 3...
a) m can be 2 and n = 3...out
b) n can be 2 and m = 3...out
c) m = 2 and n=3 ...out
d) n can be 2 and m = 3...out
e) m = 2/3 or n = 3/2 both satisfies .......must be a multiple of 9
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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19 May 2014, 21:41
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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05 Aug 2015, 23:50
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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18 Sep 2016, 23:48
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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28 Sep 2016, 07:43
Either m or n is 3,
so "E" is correct
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]

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28 Sep 2016, 12:42
A) No since n could be 9
B) No since m=3
C) No since m could be 2 and n=3
D) No since M could 3 and n=2
E) Yes since either M or N have to be 3 (or factor) so they will become a factor of 9
Re: If k is a multiple of 3 and k = (m^2)n, where m and n are   [#permalink] 28 Sep 2016, 12:42
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