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shagunnayyar
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700+ level question 27 Mar 2017, 17:54
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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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700+ level question 27 Mar 2017, 18:00
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shagunnayyar
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700+ level question 27 Mar 2017, 19:50
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Could you explain how you reached the answer. I also got the same answer but unfortunately the OA is different.
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KarishmaB
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700+ level question 27 Mar 2017, 22:49
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shagunnayyar
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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700+ level question 27 Mar 2017, 23:25
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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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700+ level question 28 Mar 2017, 07:39
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shagunnayyar
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: if-a-and-n-are-integers-and-a-3-360n-then-n-must-be-divisible-by-227801.html

Please follow the rules when posting a question. Thank you.



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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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