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

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

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New post 29 Jun 2014, 08:47
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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.
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Re: If a and n are integers, and a^2 = 24n, then n must be  [#permalink]

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

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New post 29 Jun 2014, 11:35
goodyear2013 wrote:
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.


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

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

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New post 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 ?


The question asks about divisibility of n, not a.

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).

Answer: A.

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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New post 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

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

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

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New post 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)

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

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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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