$31 \cdot 34$ satisfies the equation $\frac{a+b^{-1}}{a^{-1}+b}=13$ and $a+b \leqslant 100$ for the number of positive integer pairs $(a, b)$ is
(A) 1 .
(B) 5 .
(C) 7 .
(D) 9 .
(E) 13 .
(43rd American High School Mathematics Examination, 1992)
［Solution］ From the given equation, we have

that is,
$$
\begin{array}{c}
\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13, \\
a^{2} b+a=13\left(a b^{2}+b\right),
\end{array}
$$

or
$$
a^{2} b-13 a b^{2}+a-13 b=0 \text {. }
$$

Then $a b(a-13 b)+(a-13 b)=0$, or $(a-13 b)(a b+1)=0$.
So $a=13 b$ (noting that $a b+1 \neq 0$ ).
Also, $a+b \leqslant 100$, then $13 b+b=14 b \leqslant 100$, thus
$$
b \leqslant\left[\frac{100}{14}\right]=7 \text {. }
$$

Therefore, $b=1,2, \cdots, 7 ; a=13,26, \cdots, 91$, there are 7 pairs of $(a, b)$.
Hence, the answer is (C).