If y is an integer, is y^3 divisible by 8? : GMAT Data Sufficiency (DS)
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# If y is an integer, is y^3 divisible by 8?

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If y is an integer, is y^3 divisible by 8? [#permalink]

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16 Sep 2012, 08:45
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If y is an integer, is y^3 divisible by 8?

(1) y is even.
(2) y^3 - y is even.

[Reveal] Spoiler:
Shouldn't the answer be E because y could be 0 which is an even integer?
[Reveal] Spoiler: OA
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Re: If y is an integer, is y(cube) divisible by 8? [#permalink]

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16 Sep 2012, 08:57
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I chose A.

(1) y^3 will always be divisible by 2^3 if y is even. Ex. y =2, 4, etc

Lets say y = 0 -> even integer
-> y^3 = 0
-> y^3/2^3 = 0/8
-> 0/8 = 0 [ 0/N = 0 for any N from -infinity to +infinity, except 0.]
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Re: If y is an integer, is y(cube) divisible by 8? [#permalink]

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16 Sep 2012, 09:08
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crackmba2012 wrote:
If y is an integer, is y(cube) divisible by 8?
(1) y is even.
(2) y(cube)- y is even.

Shouldn't the answer be E because y could be 0 which is an even integer?

0 is even and it is divisible by 8. 0 is divisible by any non-zero integer.

(1) $$y=2k,$$ for some integer $$k,$$ then $$y^3=8k^3,$$ obviously divisible by 8.
Sufficient.

(2) Check for example for $$y=1$$ and $$y = 2.$$ You obtain 0 and 6. Both are even, but 6 is not divisible by 8.
Not sufficient.

Remark: $$y^3-y=y(y^2-1)=(y-1)y(y+1)$$ is the product of three consecutive integers, therefore it is always even.
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Re: If y is an integer, is y(cube) divisible by 8? [#permalink]

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16 Sep 2012, 09:11
EvaJager wrote:

Remark: $$y^3-y=y(y^2-1)=(y-1)y(y+1)$$ is the product of three consecutive integers, therefore it is always even.

Awesome! I used a very long way to get here. Should have spotted this.
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Re: If y is an integer, is y^3 divisible by 8? [#permalink]

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24 Oct 2015, 11:49
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Re: If y is an integer, is y^3 divisible by 8?   [#permalink] 24 Oct 2015, 11:49
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