One TV station has $n$ ad breaks in a day, during which a total of $m$ ads were broadcast. In the first ad break, one ad and $\frac{1}{8}$ of the remaining $(m-1)$ ads were broadcast. In the second ad break, 2 ads and $\frac{1}{8}$ of the remaining ads were broadcast. This pattern continued for each subsequent ad break, with the last ad break, the $n$-th one, broadcasting the remaining ads $(n>1)$. How many ad breaks were there that day? And what is the number of ads $m$?
Let's denote the number of ads left to be broadcast after $k$ broadcasts as $a_{k}$. Then, the number of ads broadcast at the $k$-th time is:
$$
k+\frac{1}{8}\left(a_{k-1}-k\right)=\frac{1}{8} a_{k-1}+\frac{7}{8} k .
$$

Thus, $a_{k}=a_{k-1}-\left(\frac{1}{8} a_{k-1}+\frac{7}{8} k\right)=\frac{7}{8} a_{k-1}-\frac{7}{8} k$.
$$
\therefore a_{k-1}=k+\frac{8}{7} a_{k} \text {. }
$$

From this recursive formula, we get
$$
\begin{aligned}
m= & 1+\frac{8}{7} a_{1}=1+\frac{8}{7}\left(2+\frac{8}{7} a_{2}\right) \\
= & 1+2 \times \frac{8}{7}+\left(\frac{8}{7}\right)^{2} \cdot a_{2} \\
= & 1+2 \times \frac{8}{7}+\left(\frac{8}{7}\right)^{2} \cdot\left(3+\frac{8}{7} a_{3}\right) \\
= & 1+2 \times \frac{8}{7}+3 \times\left(\frac{8}{7}\right)^{2}+\left(\frac{8}{7}\right)^{3} \times a_{3} \\
= & \cdots \cdots \\
= & 1+2 \times \frac{8}{7}+3 \times\left(\frac{8}{7}\right)^{2}+\cdots+n \times\left(\frac{8}{7}\right)^{n-1} \\
& +\left(\frac{8}{7}\right)^{n} \cdot a_{n} .
\end{aligned}
$$

By the definition of $a_{k}$, we know that $a_{n}=0$. Therefore,
$$
m=\sum_{i=1}^{n} i\left(\frac{8}{7}\right)^{i-1} \text {. }
$$
$$
\therefore \frac{8}{7} m=\sum_{i=1}^{n} i\left(\frac{8}{7}\right)^{i} \text {. }
$$
Subtracting (1) from (2) gives
$$
\begin{array}{l}
-\frac{1}{7} m= 1+\frac{8}{7}+\left(\frac{8}{7}\right)^{2}+\cdots+\left(\frac{8}{7}\right)^{n-1} \\
-n \cdot\left(\frac{8}{7}\right)^{n} \\
= \frac{\left(\frac{8}{7}\right)^{n}-1}{\frac{8}{7}-1}-n \cdot\left(\frac{8}{7}\right)^{n} . \\
\therefore m=49+(n-7) \cdot \frac{8^{n}}{7^{n-1}} .
\end{array}
$$

For $n>1$, it is clear that
$$
|n-7|<7^{n-1},\left(8^{n}, 7^{n-1}\right)=1 \text {. }
$$

Since $m$ is an integer, it must be that $n-7=0$, i.e., $n=7$. Thus, $m=49$.
Therefore, there are 7 broadcasts, and a total of 1,9 ads are broadcast.