33. (1988 American Mathematical Invitational) If a positive divisor of $10^{99}$ is chosen at random, the probability that it is exactly a multiple of $10^{88}$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime, find $m+n$.
33. The divisors of $10^{99}$ are all of the form $2^{a} \cdot 5^{b}$, where $a$ and $b$ satisfy $0 \leqslant a \leqslant 99,0 \leqslant b \leqslant 99$.

Therefore, $10^{99}$ has $(99+1)(99+1)=100 \times 100$ positive divisors.
Among these positive divisors, to be a multiple of $10^{88}=2^{88} \cdot 5^{88}$, they must satisfy
$$
88 \leqslant a \leqslant 99,88 \leqslant b \leqslant 99 \text {. }
$$

That is, $a$ and $b$ each have 12 choices.
Thus, the required probability is: $\frac{m}{n}=\frac{12 \times 12}{100 \times 100}=\frac{9}{625}$.

Therefore, $m+n=625+9=634$.