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# X is divisible by 144. If X^1/3 is an integer, then which of

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Joined: 22 Mar 2010
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Schools: HEC Paris, Wharton,UMD
X is divisible by 144. If X^1/3 is an integer, then which of [#permalink]

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30 Mar 2010, 17:48
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X is divisible by 144. If X^1/3 is an integer, then which of the following is X^1/3 definitely divisible by? (choose all that apply)
a) 4
b) 8
c) 9
d) 12
Note : X^1/3 is the cubic root of X.

This question is from ManhattanGMAT, they provided an answer and explanation which I don't understand.

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Joined: 21 Jul 2009
Posts: 364
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

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30 Mar 2010, 18:05
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lylymenko wrote:
This question is from ManhattanGMAT, they provided an answer and explanation which I don't understand.

X is divisible by 144. If X^1/3 is an integer, then which of the following is X^1/3 definitely divisible by? (choose all that apply)
a) 4
b) 8
c) 9
d) 12
Note : X^1/3 is the cubic root of X.

My take are both A and D.

X is divisible by 144 = 12 squared = 3 * 4 * 3 * 4.
For the cubic root of x to be an integer, consider multiplying a value y to 144 to get x such that cubic root of 3*4*3*4*y is an integer. Basic exponents tells us that until y is another 3*4, it is practically not possible to get a cubic root which is also an integer. Thus after calculating the cube root what remains is a 3*4, certainly divisible by both A and D options given.
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Math Expert
Joined: 02 Sep 2009
Posts: 39698

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30 Mar 2010, 18:16
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lylymenko wrote:
This question is from ManhattanGMAT, they provided an answer and explanation which I don't understand.

X is divisible by 144. If X^1/3 is an integer, then which of the following is X^1/3 definitely divisible by? (choose all that apply)
a) 4
b) 8
c) 9
d) 12
Note : X^1/3 is the cubic root of X.

$$x=144k=2^4*3^2k$$
$$\sqrt[3]{x}=\sqrt[3]{144k}=\sqrt[3]{2^4*3^2k}=2\sqrt[3]{2*3^2k}=integer$$ --> as $$2\sqrt[3]{2*3^2k}$$ is an integer, then min value of $$k$$ is $$2^2*3$$ (to complete the powers of 2 and 3 under the cubic root). So min value of $$\sqrt[3]{x}$$ is $$\sqrt[3]{x}=2\sqrt[3]{2*3^2k}=2\sqrt[3]{2*3^2*2^2*3}=12$$. 12 is divisible by 4 and 12.

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Joined: 22 Mar 2010
Posts: 4
Schools: HEC Paris, Wharton,UMD

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30 Mar 2010, 18:55
Its all make a lot of sense now, thanks to you all
Re: Can someone answer this and tell me why?   [#permalink] 30 Mar 2010, 18:55
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