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

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Intern
Joined: 07 Sep 2012
Posts: 34
GMAT 1: 680 Q50 V32
GMAT 2: 700 Q49 V36
If y is an integer, is y^3 divisible by 8?  [#permalink]

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16 Sep 2012, 09:45
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Difficulty:

15% (low)

Question Stats:

78% (01:18) correct 22% (01:18) wrong based on 195 sessions

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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.

Shouldn't the answer be E because y could be 0 which is an even integer?
Senior Manager
Joined: 28 Jun 2009
Posts: 372
Location: United States (MA)
Re: If y is an integer, is y(cube) divisible by 8?  [#permalink]

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16 Sep 2012, 09:57
1
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.]
Director
Joined: 22 Mar 2011
Posts: 599
WE: Science (Education)
Re: If y is an integer, is y(cube) divisible by 8?  [#permalink]

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16 Sep 2012, 10:08
2
1
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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Senior Manager
Joined: 28 Jun 2009
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Location: United States (MA)
Re: If y is an integer, is y(cube) divisible by 8?  [#permalink]

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16 Sep 2012, 10:11
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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17 Jul 2017, 19:25
crackmba2012 wrote:
If y is an integer, is y^3 divisible by 8?

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

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

St 1

Is clearly suff - for example 6^ 3

3 x 2 x 3 x 2 x 3 x2/ 2 * 2 *2

St 2

(5)^3-5= 120 ... but wait...we're only evaluating for y^3 so insuff

A
Re: If y is an integer, is y^3 divisible by 8?   [#permalink] 17 Jul 2017, 19:25
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