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# If y is an integer, is y^3 divisible by 12?

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Math Expert
Joined: 02 Sep 2009
Posts: 56244
If y is an integer, is y^3 divisible by 12?  [#permalink]

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10 Dec 2017, 01:20
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25% (medium)

Question Stats:

88% (01:03) correct 12% (01:57) wrong based on 59 sessions

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If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.

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Joined: 25 Feb 2013
Posts: 1197
Location: India
GPA: 3.82
If y is an integer, is y^3 divisible by 12?  [#permalink]

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10 Dec 2017, 03:16
1
Bunuel wrote:
If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.

For $$y^3$$ to be divisible by $$12$$, $$y$$ must contain at least one $$2$$ & one $$3$$.

Statement 1: $$y =4p =>y^3=64p^3$$. $$p$$ can be a multiple of $$3$$ or not. Hence Insufficient

Statement 2: $$y=6k =>y^3=(2*3)^3k^3$$. Clearly divisible by $$12$$. Sufficient

Option B
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Joined: 04 Aug 2017
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If y is an integer, is y^3 divisible by 12?  [#permalink]

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24 Dec 2017, 05:36
niks18 wrote:
Bunuel wrote:
If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.

For $$y^3$$ to be divisible by $$12$$, $$y$$ must contain at least one $$2$$ & one $$3$$.

Statement 1: $$y =4p =>y^3=64p^3$$. $$p$$ can be a multiple of $$3$$ or not. Hence Insufficient

Statement 2: $$y=6k =>y^3=(2*3)^3k^3$$. Clearly divisible by $$12$$. Sufficient

Option B

Would you be able to explain why y has to have least one $$2$$ & one $$3$$ and not two $$2s$$ & one $$3$$ Thank you!
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Re: If y is an integer, is y^3 divisible by 12?  [#permalink]

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24 Dec 2017, 05:44
imranmah wrote:
niks18 wrote:
Bunuel wrote:
If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.

For $$y^3$$ to be divisible by $$12$$, $$y$$ must contain at least one $$2$$ & one $$3$$.

Statement 1: $$y =4p =>y^3=64p^3$$. $$p$$ can be a multiple of $$3$$ or not. Hence Insufficient

Statement 2: $$y=6k =>y^3=(2*3)^3k^3$$. Clearly divisible by $$12$$. Sufficient

Option B

Would you be able to explain why y has to have least one $$2$$ & one $$3$$ and not two $$2s$$ & one $$3$$ Thank you!

Hi imranmah

y can have more than one 2 and 3 but it has to have AT LEAST one 2 & 3.

so let's assume $$y=2*3=6$$ then $$y^3=6*6*6$$. therefore $$\frac{y^3}{12}=\frac{6*6*6}{12}$$, divisible by 12

and as per your example if $$y=2*2*3$$, then $$\frac{y^3}{12}=\frac{2*2*2*2*2*2*3*3*3}{12}$$, divisible by 12

but it cannot be that y has only one 2 and no 3 or vice-versa.
Re: If y is an integer, is y^3 divisible by 12?   [#permalink] 24 Dec 2017, 05:44
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