How many different pairs of positive integers $(a, b)$ are there such that $a+b \leq 100$ and $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$?
(a) 1
(b) 5
(c) 7
(d) 9
(e) 13
The correct option is (c).

We have

$$
13=\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=\frac{\frac{a b+1}{b}}{\frac{1+a b}{a}}=\frac{(a b+1) \times a}{(1+a b) \times b}=\frac{a}{b}
$$

Thus, $a=13 b$. Since $a+b \leq 100$, it follows that $14 b \leq 100$ and, therefore, $b \leq 7.14$. Since $b$ is an integer, we must have $b \leq 7$. Therefore, the pairs are seven in number, namely,

$$
(13,1),(26,2),(39,3),(52,4),(65,5),(78,6) \text { and }(91,7)
$$