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Re: GrockIt: divisibility [#permalink]
B for me. I believe this problem can also be solved by using prime boxes.
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Re: GrockIt: divisibility [#permalink]
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]
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]
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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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
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If y is an integer greater than 0, is (y^3 - y) divisible by 4? [#permalink]
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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HKD1710 wrote:
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


Merging topics. Please refer to the discussion above.
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Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]
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 formating for formulas.


Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 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 y-1 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]
coolkl wrote:
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 formating for formulas.



Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 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 y-1 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 (y-1)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]
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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TheMastermind wrote:
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.


You should brush up fundamentals.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: https://gmatclub.com/forum/number-proper ... 74996.html
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Re: If y > 0, is (y^3 y) divisible by 4? [#permalink]
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Re: If y > 0, is (y^3 y) divisible by 4? [#permalink]
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