If $a=7$ and $b=13$, the number of even positive integers less than $a b$ is
(A) $\frac{a b-1}{2}$
(B) $\frac{a b}{2}$
(C) $a b-1$
(D) $\frac{a+b}{4}$
(E) $(a-1)(b-1)$
Solution 1

Since $a$ and $b$ are both odd, then $a b$ is odd.

Therefore, the largest even integer less than $a b$ is $a b-1$.

Since every other positive integer less than or equal to $a b-1$ is even, then the number of even positive integers less than or equal to $a b-1$ (thus, less than $a b$ ) is $\frac{a b-1}{2}$.

## Solution 2

Since $a=7$ and $b=13$, then $a b=91$.

The even positive integers less than $a b=91$ are $2,4,6, \ldots, 90$.

There are $90 \div 2=45$ such integers.

Using $a=7$ and $b=13$, the five possible answers are

$$
\frac{a b-1}{2}=45 \quad \frac{a b}{2}=\frac{91}{2} \quad a b-1=90 \quad \frac{a+b}{4}=5 \quad(a-1)(b-1)=72
$$

Therefore, the answer must be $\frac{a b-1}{2}$.

ANSWER: (A)