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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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New post 16 Sep 2012, 09:45
2
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A
B
C
D
E

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  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?
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Re: If y is an integer, is y(cube) divisible by 8?  [#permalink]

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New post 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
(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]

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New post 16 Sep 2012, 10:08
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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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Re: If y is an integer, is y(cube) divisible by 8?  [#permalink]

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