7. 8 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 coprime, find $m+n$.
［Solution］The divisors of $10^{99}$ all have 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 $10^{88}=2^{88} \cdot 5^{88}$, it must satisfy
$$88 \leqslant a \leqslant 99,88 \leqslant b \leqslant 99$$

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

So $m+n=625+9=634$.