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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- Stne