$32 \cdot 30$ is greater than 10, less than 100, when written in decimal, the number of integers whose digits, when reversed, are 9 more than the original number is
(A) 0.
(B) 1.
(C) 8.
(D) 9.
(E) 10.
(27th American High School Mathematics Examination, 1976)
[Solution] Let $t$ and $u$ be the tens and units digits of this number, respectively. Given that when the digits are swapped, the new number is 9 more than the original number, we have
$$
9=(10 u+t)-(10 t+u)=9(u-t),
$$

i.e.,
$$
u=t+1 \text {. }
$$

Also, $0 \leqslant t \leqslant 9,1 \leqslant u \leqslant 9$, thus we get 8 solutions, which are $\{12,23, \cdots, 89\}$.

Therefore, the answer is $(C)$.