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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # If n is an integer and n3 is divisible by 24, what is the la  Question banks Downloads My Bookmarks Reviews Important topics
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Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 175

Kudos [?]: 3033 , given: 394

If n is an integer and n3 is divisible by 24, what is the la [#permalink]
Expert's post 00:00

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

_________________

Sandy
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Try our free Online GRE Test Intern Joined: 02 Aug 2018
Posts: 1
Followers: 0

Kudos [?]: 2  , given: 1

Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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: 175

Kudos [?]: 3033  , given: 394

Re: If n is an integer and n3 is divisible by 24, what is the la [#permalink]
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 $$2 \times 3 = 6$$.
_________________

Sandy
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Try our free Online GRE Test 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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