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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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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.
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Re: GrockIt: divisibility [#permalink]
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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^3y divisible by 4? \(y^3y=y(y^21)=y(y1)(y+1)\) (1) y^2+y is divisible by 10 > if \(y=10\) then \(y(y1)(y+1)=10*9*11\), 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 \(y(y1)(y+1)=100*99*101\), 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 > \(y=odd\) > \(y1=even\) and \(y+1=even\) > \(y(y1)(y+1)\) is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as \(y1\) and \(y1\) 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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26 Aug 2010, 12:18
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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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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Re: GrockIt: divisibility [#permalink]
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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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Re: GrockIt: divisibility [#permalink]
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03 Mar 2014, 04:57
Bunuel wrote: If y>0, is y^3y divisible by 4?
\(y^3y=y(y^21)=y(y1)(y+1)\)
(1) y^2+y is divisible by 10 > if \(y=10\) then \(y(y1)(y+1)=10*9*11\), 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 \(y(y1)(y+1)=100*99*101\), 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 > \(y=odd\) > \(y1=even\) and \(y+1=even\) > \(y(y1)(y+1)\) is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as \(y1\) and \(y1\) 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 (y1) and (y+1) are 2 consecutive even numbers and not (y1 ) and ( y1) 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
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Bunuel wrote: If y>0, is y^3y divisible by 4?
\(y^3y=y(y^21)=y(y1)(y+1)\)
(1) y^2+y is divisible by 10 > if \(y=10\) then \(y(y1)(y+1)=10*9*11\), 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 \(y(y1)(y+1)=100*99*101\), 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 > \(y=odd\) > \(y1=even\) and \(y+1=even\) > \(y(y1)(y+1)\) is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as \(y1\) and \(y1\) 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



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If y is an integer greater than 0, is (y^3  y) divisible by 4? [#permalink]
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15 Jan 2017, 08:20
If \(y\) is an integer greater than 0, is (\(y^3\)  \(y\)) divisible by 4? (1) \(y^2\) + \(y\) is divisible by 10. (2) For a certain integer \(k\), \(y\) = 2\(k\) + 1
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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15 Jan 2017, 08:34



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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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09 Mar 2017, 11:04
Bunuel wrote: If y>0, is y^3y divisible by 4?
\(y^3y=y(y^21)=y(y1)(y+1)\)
(1) y^2+y is divisible by 10 > if \(y=10\) then \(y(y1)(y+1)=10*9*11\), 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 \(y(y1)(y+1)=100*99*101\), 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 > \(y=odd\) > \(y1=even\) and \(y+1=even\) > \(y(y1)(y+1)\) is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as \(y1\) and \(y1\) are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.
Answer: B.
Pleas use "^" for powers, and formating for formulas. Hello Bunnel or GMATCLUB members, For Statement 2 we say since y is odd means y1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0. Now y1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help?
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If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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09 Mar 2017, 11:07
coolkl wrote: Bunuel wrote: If y>0, is y^3y divisible by 4?
\(y^3y=y(y^21)=y(y1)(y+1)\)
(1) y^2+y is divisible by 10 > if \(y=10\) then \(y(y1)(y+1)=10*9*11\), 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 \(y(y1)(y+1)=100*99*101\), 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 > \(y=odd\) > \(y1=even\) and \(y+1=even\) > \(y(y1)(y+1)\) is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as \(y1\) and \(y1\) are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.
Answer: B.
Pleas use "^" for powers, and formating for formulas. Hello Bunnel or GMATCLUB members, For Statement 2 we say since y is odd means y1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0. Now y1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help? I got my answer, if y =1 then (y1)y(y+1) is 0 and 0 div by 4 is still 0. This is good question.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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30 May 2017, 06:15
I got to understand that 'y' is odd as given in statement (2) but if y=1, the expression given would become zero. So it shouldn't be considered as divisible by 4.



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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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30 May 2017, 07:57




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