Interesting question. Kudos to you. Very simple explanation though

Question: If \(y\) is a positive integer , is \(y^2 - y\) divisible by \(4\)? Lets factor \(y^2 - y\):

\(y^2-y=y*(y-1)\)

So we know that these are two consecutive integers, so one of them has to be odd and one of them is even. Which means only one of them can be a multiple of \(4\). So we need to establish that either \(y\) is a multiple of \(4\) or \((y-1)\) is a multiple of \(4\).

Statement A: \(y^2 + y\) is not divisible by \(4\)

Lets factor this one out : \(y^2+y=y*(y+1)\)

These again are consecutive integers and the statement says that niether of them is a factor of \(4\). Remember our original conditions. either \(y\) is a multiple of \(4\) or \((y-1)\) is a multiple of \(4\). This statement tells us that \(y\) is not a multiple of \(4\) but it does not tell us whether \((y-1)\) is a multiple of \(4\) or not. Also note that this again does not tell us whether \(y\) is even or \((y-1)\) is even.

So Insufficient.

Statement B: \(y^3-y\) is divisible by \(4\)

Lets factor this one out : \(y^3-y=y*(y-1)*(y+1)\) and again these are \(3\) consecutive integers. The only thing we have found out (from the statement) is that one of them is definitely a multiple of \(4\) but we do not know which one it is. Also note that we are not sure that one of them is even or two of them are even. And if it is one even, it could be \(2\) so it is still not sufficient. We need at-least \(2\) multiples of \(2\) in there. If there were \(2\) even numbers, we would be sure that it is divisible by \(4\).

So InsufficientCombined:

We know that \(y\) and \((y+1)\) are not multiples of \(4\) from the first statement so from our second statement combined we have found out that \((y-1)\) is then, logically speaking, a multiple of \(4\) and since \((y-1)\) is one of our original factors from the question stem, the answer is YES. \(y^2-1\) whose factors are \(y*(y-1)\) is divisible by \(4\) because of the presence of \((y-1)\) as a factor. A Kudos won't hurt here!

Answer

C
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