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700+ level question

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700+ level question [#permalink]

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New post 27 Mar 2017, 16:54
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If a and n are integers, and a3=360n, then n must be divisible by which of the following?

1) 2
2) 6
3) 25
4) 27
5) 60
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Re: 700+ level question [#permalink]

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New post 27 Mar 2017, 17:00
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Re: 700+ level question [#permalink]

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New post 27 Mar 2017, 18:50
Could you explain how you reached the answer. I also got the same answer but unfortunately the OA is different.
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Re: 700+ level question [#permalink]

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New post 27 Mar 2017, 21:49
shagunnayyar wrote:
If a and n are integers, and a3=360n, then n must be divisible by which of the following?

1) 2
2) 6
3) 25
4) 27
5) 60


I am assuming you mean
\(a^3 = 360n\)

\(a^3 = 2*2*2*3*3*5*n\)

\(a^3 = 2^3 * 3^2 * 5 * n\)

Considering the cube of an integer, every prime factor will have an exponent that is a multiple of 3.
So n must have a 3 and two 5s at least so that you get \(2^3 * 3^3 * 5^3\) (every prime will have an exponent that is a power of 3).
Note that n could be \(3*5^5\) or \(3^4*5^2\) etc (infinite possibilities)

So n must be divisible by 3*5*5.
Of the given options, n must be divisible by 5*5 = 25

Answer (C)
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Re: 700+ level question [#permalink]

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New post 27 Mar 2017, 22:25
Quote:
If a and n are integers, and a3=360n, then n must be divisible by which of the following?

1) 2
2) 6
3) 25
4) 27
5) 60


Hi shagunnayyar
Please use math mode for writing equations. It should be \(a^{3} = 360n\).

Solution:

\(360n = 2^{3}\times 3^{2} \times 5 \times n\)

To a perfect cube, the minimum value of \(n\) should be 3*5*5.

=> \(n\) must be divisible by 25.

Hope this helps.
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Re: 700+ level question [#permalink]

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New post 28 Mar 2017, 06:39
shagunnayyar wrote:
If a and n are integers, and a3=360n, then n must be divisible by which of the following?

1) 2
2) 6
3) 25
4) 27
5) 60


This question is discussed here: https://gmatclub.com/forum/if-a-and-n-a ... 27801.html

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Re: 700+ level question   [#permalink] 28 Mar 2017, 06:39
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