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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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Manager
Joined: 28 May 2009
Posts: 144
Location: United States
Concentration: Strategy, General Management
GMAT Date: 03-22-2013
GPA: 3.57
WE: Information Technology (Consulting)
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, 12:34
3
9
00:00

Difficulty:

35% (medium)

Question Stats:

70% (01:32) correct 30% (01:42) wrong based on 393 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

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Joined: 16 Oct 2010
Posts: 8787
Location: Pune, India
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, 20:19
6
1
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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Karishma
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##### General Discussion
Manager
Joined: 18 Oct 2011
Posts: 86
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
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, 13:05
1
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.
Manager
Joined: 27 Feb 2012
Posts: 119
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, 13:28
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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Please +1 KUDO if my post helps. Thank you.

Math Expert
Joined: 02 Sep 2009
Posts: 52161
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, 02:01
1
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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Intern
Joined: 20 Sep 2016
Posts: 1
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, 06:43
Either m or n is 3,
so "E" is correct
Intern
Joined: 23 Jun 2016
Posts: 41
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, 11:42
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
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Joined: 09 Sep 2013
Posts: 9415
Re: If k is a multiple of 3 and k = (m^2)n, where m and n are  [#permalink]

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09 Feb 2018, 13:00
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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are &nbs [#permalink] 09 Feb 2018, 13:00
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