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# If y > 0, is (y^3 – y) divisible by 4?

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Manager
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If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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26 Aug 2010, 05:31
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60% (02:34) correct 40% (02:07) wrong based on 124 sessions

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If y > 0, is (y^3 – y) divisible by 4?

(1) y^2 + y is divisible by 10.
(2) For a certain integer k, y = 2k + 1.
[Reveal] Spoiler: OA
Manager
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26 Aug 2010, 05:34
zisis wrote:
If y > 0, is (y3 – y) divisible by 4?

(1) y2 + y is divisible by 10.

(2) For a certain integer k, y = 2k + 1.

my method

$$1. y(y+1)$$ is div by 10

therefore if y is div by 10 (y^3 -y) is div by 4 BUT if y+1 is div by 10 therefore y= 9, 19 etc thus y+1 is not div by 4

2. y is odd therefore is not div by 4

any better methods?
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If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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26 Aug 2010, 07:32
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If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.
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26 Aug 2010, 12:18
B for me. I believe this problem can also be solved by using prime boxes.
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26 Aug 2010, 12:30
the key point to this question is to find out if y is an odd#...i pick B
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27 Aug 2010, 10:03
I pick B and I agree with tt11234...The key here is to pick that y is odd!
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03 Mar 2014, 04:57
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

answer is definitely B , as the product of 2 consecutive even numbers is divisible by 4.
There is a small typo above , it should be (y-1) and (y+1) are 2 consecutive even numbers and not (y-1 ) and ( y-1) are 2 consecutive even numbers ,

Hope this prevents any confusion.
Thank you.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

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10 Nov 2015, 09:02
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formatting for formulas.

I think one more case is to be considered, i.e when k = 0, y = 1. Even then 0/4 = 0 a integer. B will remain the answer
Re: If y > 0, is (y^3 – y) divisible by 4?   [#permalink] 10 Nov 2015, 09:02
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