If $a$ is an even integer and $b$ is an odd integer, which of the following could represent an odd integer?
(A) $a b$
(B) $a+2 b$
(C) $2 a-2 b$
(D) $a+b+1$
(E) $a-b$
Solution 1

In this solution, we try specific values. This does not guarantee correctness, but it will tell us which answers are wrong.

We try setting $a=2$ (which is even) and $b=1$ (which is odd).

Then $a b=2 \times 1=2, a+2 b=2+2(1)=4,2 a-2 b=2(2)-2(1)=2, a+b+1=2+1+1=4$, and $a-b=2-1=1$.

Therefore, $a-b$ is the only choice which gives an odd answer.

## Solution 2

Since $a$ is even, then $a b$ is even, since an even integer times any integer is even.

Since $a$ is even and $2 b$ is even (since 2 times any integer is even), then their sum $a+2 b$ is even. Since 2 times any integer is even, then both $2 a$ and $2 b$ are both even, so their difference $2 a-2 b$
is even (since even minus even is even).

Since $a$ is even and $b$ is odd, then $a+b$ is odd, so $a+b+1$ is even.

Since $a$ is even and $b$ is odd, then $a-b$ is odd.

Therefore, $a-b$ is the only choice which gives an odd answer.

ANSWER: (E)