$7 \cdot 8$ If a positive divisor is randomly selected from $10^{99}$, the probability that it is exactly a multiple of $10^{88}$ is $\frac{m}{n}$, where $m$ and $n$ are coprime, find $m+n$.

The prime factorization of $10^{99}$ is $2^{99} \cdot 5^{99}$. The total number of positive divisors of $10^{99}$ is $(99+1)(99+1) = 100 \cdot 100 = 10000$.

For a divisor to be a multiple of $10^{88}$, it must be of the form $2^{a} \cdot 5^{b}$ where $a \geq 88$ and $b \geq 88$. The number of such divisors is $(100-88)(100-88) = 12 \cdot 12 = 144$.

Thus, the probability that a randomly selected divisor of $10^{99}$ is a multiple of $10^{88}$ is $\frac{144}{10000} = \frac{9}{625}$. Here, $m = 9$ and $n = 625$, and they are coprime.

Therefore, $m+n = 9 + 625 = 634$.
[Solution] 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
$$
88 \leqslant a \leqslant 99, 88 \leqslant b \leqslant 99 .
$$

That is, $a$ 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 $m+n=625+9=634$.