If y is an integer, is y(cube) divisible by 8?
(1) y is even.
(2) y(cube)- y is even.
Shouldn't the answer be E because y could be 0 which is an even integer?
0 is even and it is divisible by 8. 0 is divisible by any non-zero integer.
(1) \(y=2k,\) for some integer \(k,\) then \(y^3=8k^3,\) obviously divisible by 8.
(2) Check for example for \(y=1\) and \(y = 2.\) You obtain 0 and 6. Both are even, but 6 is not divisible by 8.
Remark: \(y^3-y=y(y^2-1)=(y-1)y(y+1)\) is the product of three consecutive integers, therefore it is always even.
PhD in Applied Mathematics
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