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# If n is an integer and n3 is divisible by 24, what is the la

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Retired Moderator
Joined: 07 Jun 2014
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GRE 1: Q167 V156
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If n is an integer and n3 is divisible by 24, what is the la [#permalink]  12 Aug 2018, 15:46
Expert's post
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Question Stats:

38% (01:06) correct 61% (01:18) wrong based on 36 sessions
If n is an integer and $$n^3$$ is divisible by 24, what is the largest number that must be a factor of n?

(A) 1
(B) 2
(C) 6
(D) 8
(E) 12
[Reveal] Spoiler: OA

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Sandy
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Intern
Joined: 02 Aug 2018
Posts: 1
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Kudos [?]: 2 [1] , given: 1

Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]  13 Aug 2018, 00:11
1
KUDOS
why 6 but not 12?
Retired Moderator
Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 171

Kudos [?]: 2915 [1] , given: 394

Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]  17 Aug 2018, 16:08
1
KUDOS
Expert's post
Explanation

Start by considering the relationship between $$n$$ and $$n^3$$. Because $$n$$ is an integer, for every prime factor $$n$$ has, $$n^3$$ must have three of them.

Thus, $$n^3$$ must have prime numbers in multiples of 3. If $$n^3$$ has one prime factor of 3, it must actually have two more, because $$n^3$$’s prime factors can only
come in triples.

The question says that $$n^3$$ is divisible by 24, so $$n^3$$’s prime factors must include at least three 2’s and a 3.

But since $$n^3$$ is a cube, it must contain at least three 3’s. Therefore, n must contain at least one 2 and one 3, or $$2 \times 3 = 6$$.
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Sandy
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Re: If n is an integer and n3 is divisible by 24, what is the la   [#permalink] 17 Aug 2018, 16:08
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