, 2015

When $n=7$, we have $$9n=63 \qquad n+8 = 15 \qquad n^2 = 49 \qquad n(n-2) = 7(5)=35 \qquad 8n = 56$$ Therefore, $8n$ is even.

We note that for every integer $n$, the expression $8n$ is equal to an even integer, since 8 is even and the product of an even integer with any integer is even.

If $n$ were even, then in fact all five choices would be even. If $n$ is odd, only $8n$ is even.