Since this is a divisibility question, we should certainly try to get factorizations. The expression y^3 - y can be factored first by factoring out one y, and then by using the difference of squares:dell2 wrote:If y is an integer greater than 0, is (y3 - y) divisible by 4?
(1) y2 + y is divisible by 10.
(2) For a certain integer k, y = 2k + 1.
y^3 - y = y(y^2 - 1) = (y)(y-1)(y+1)
Now (y-1)(y)(y+1) is just the product of three consecutive integers. If, as Statement 2 tells us, y is odd, then y-1 and y+1 are both even, so their product must be divisible by 2^2. So Statement 2 is sufficient.
From Statement 1 we know that (y)(y+1) is divisible by 2*5. That will certainly be true if y = 10, but if y = 10, then (y-1)(y)(y+1) = 9*10*11 is *not* divisible by 4. So with Statement 1 alone, the answer to the question can be 'no'. It can also be 'yes' of course (let y = 5, say, or any other odd multiple of 5), so Statement 1 is not sufficient, and the answer is B.
