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TAGS: GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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S be the set of all positive integers n such that n^2 [#permalink]
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Expert's post 00:00

Question Stats: 43% (01:10) correct 56% (01:59) wrong based on 16 sessions
Let S be the set of all positive integers n such that $$n^2$$ is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S ?

Indicate all such integers.

A. 12
B. 24
C. 36
D. 72

Practice Questions
Question: 14
Page: 159
Difficulty: hard

[Reveal] Spoiler: OA
A,C

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Sandy
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4809
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 135

Kudos [?]: 2083 , given: 397

Re: S be the set of all positive integers n such that n^2 [#permalink]
Expert's post
Now here we are looking for a Set S of numbers n such that $$n^2$$ is multiple of 24 and 108.

Lets take one such case Least Common Multiple of 108 and 24 is 216.

$$n^2$$ = 216
n = 6$$\sqrt{6}$$.

This is not an integer however if we multiply 216 by any number it is still divisible by 24 and 108. So we multiply by 6 and repat the above steps to find n. So n =36.

Now n= 36 is one of the number in set S.

Hence all the options that divide n =36 are correct. Hence option A and C.
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Intern Joined: 28 Mar 2016
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Re: S be the set of all positive integers n such that n^2 [#permalink]
can you please show the exact repeated steps please?
Manager Joined: 23 Jan 2016
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Re: S be the set of all positive integers n such that n^2 [#permalink]
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct. Re: S be the set of all positive integers n such that n^2   [#permalink] 15 May 2016, 21:09
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