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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] New post 16 Sep 2012, 08:45
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Question Stats:

79% (01:43) correct 21% (00:40) wrong based on 53 sessions
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
Senior Manager
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Re: If y is an integer, is y(cube) divisible by 8? [#permalink] New post 16 Sep 2012, 08:57
I chose A.

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

About your question,

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] New post 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.

Answer A

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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PhD in Applied Mathematics
Love GMAT Quant questions and running.

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Re: If y is an integer, is y(cube) divisible by 8? [#permalink] New post 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.
Re: If y is an integer, is y(cube) divisible by 8?   [#permalink] 16 Sep 2012, 09:11
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