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# If a and n are integers, and a^2 = 24n, then n must be

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Senior Manager
Joined: 21 Oct 2013
Posts: 419
If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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29 Jun 2014, 08:47
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Question Stats:

68% (01:09) correct 32% (01:22) wrong based on 223 sessions

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If a and n are integers, and a^2 = 24n, then n must be divisible by which of the following?

A. 2
B. 4
C. 12
D. 18
E. 24

OE
$$a^2$$=24n = 2·2·2·3·n to 2·2·2·3·2·3.
Intern
Joined: 28 Mar 2014
Posts: 21
Location: India
GPA: 3
Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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29 Jun 2014, 09:06
a^2 = 24 n = 2^2 * 2 *3 * n. Since a & n are integers n should must contain at least one 2 and one 3. Option choice i) gives 2. Therefore i) is correct
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Joined: 02 Sep 2009
Posts: 52296
Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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29 Jun 2014, 11:35
Manager
Joined: 07 Dec 2009
Posts: 91
GMAT Date: 12-03-2014
If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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07 Jul 2014, 13:47
hi,

shouldn't the answer be E ?

since a is an integer we need atleast one more 2 and 3. only with option E we get a= Integer.

With option a we will have a^2 = (2^3)*3*2. this way a is not an Integer... Where am I going wrong ?
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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07 Jul 2014, 14:10
1
bhatiavai wrote:
hi,

shouldn't the answer be E ?

since a is an integer we need atleast one more 2 and 3. only with option E we get a= Integer.

With option a we will have a^2 = (2^3)*3*2. this way a is not an Integer... Where am I going wrong ?

If a and n are integers, and a^2 = 24n, then n must be divisible by which of the following?

A. 2
B. 4
C. 12
D. 18
E. 24

$$a^2 = 24n=4*6n$$;
$$a = 2*\sqrt{6n}$$.

From above the least positive value of n is 6 and if it is 6, then it's divisible only by option A (2).

Hope it's clear.
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Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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07 Jul 2014, 20:35
$$a^2 = 24n$$

$$a^2 = 2^2 * 6 n$$

Least value of n to make "a" a perfect square is 6

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Intern
Joined: 30 Oct 2015
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GPA: 3.89
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Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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14 Nov 2015, 10:12
The biggest hint here but just be the answer choices. All of the answers are even, and thus also divisible by 2. Choice A is the only exclusive answer.
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Joined: 05 Sep 2016
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Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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30 Nov 2016, 08:33
If we break down 24 to show its prime factorization --> (2^3)(3) --> We see that we are missing 2 and 3 in order to great a perfect square.

2x3=6 = n

Since a is an integer, 6 is only divisible by 2 (it will give you an integer, not a fraction)

A.
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Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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31 Oct 2018, 22:06
Hello from the GMAT Club BumpBot!

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Re: If a and n are integers, and a^2 = 24n, then n must be &nbs [#permalink] 31 Oct 2018, 22:06
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