If y > 0, is (y^3 – y) divisible by 4? : GMAT Data Sufficiency (DS)
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If y > 0, is (y^3 – y) divisible by 4?

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

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New post 26 Aug 2010, 04:31
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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
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Re: GrockIt: divisibility [#permalink]

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New post 26 Aug 2010, 04:34
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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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.

Answer: B.

Pleas use "^" for powers, and formating for formulas.
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Re: GrockIt: divisibility [#permalink]

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New post 26 Aug 2010, 11:18
B for me. I believe this problem can also be solved by using prime boxes.
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Re: GrockIt: divisibility [#permalink]

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New post 26 Aug 2010, 11:30
the key point to this question is to find out if y is an odd#...i pick B
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Re: GrockIt: divisibility [#permalink]

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New post 27 Aug 2010, 09:03
I pick B and I agree with tt11234...The key here is to pick that y is odd!
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Re: GrockIt: divisibility [#permalink]

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New post 03 Mar 2014, 03:57
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.

Answer: B.

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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New post 10 Nov 2015, 08:02
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.

Answer: B.

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, 08:02
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