Example 2 The sports meet lasted for $n$ days $(n>1)$, and a total of $m$ medals were awarded. On the first day, 1 medal plus $\frac{1}{7}$ of the remaining medals were awarded, on the second day, 2 medals plus $\frac{1}{7}$ of the remaining medals were awarded, and so on. On the last day, the $n$-th day, exactly $n$ medals were awarded, with no medals left. How many days did the sports meet last? How many medals were awarded in total?
(9th IMO Problem)
Solution 1 Let the number of medals issued on the $k$-th day be $a_{k}$, then
$$
\begin{array}{l}
a_{1}=1+\frac{1}{7}(m-1)=\frac{1}{7}(m+6), \\
a_{k}=k+\frac{1}{7}\left(m-a_{1}-a_{2}-\cdots-a_{k-1}-k\right), \\
\quad a_{k+1}=k+1+\frac{1}{7}\left(m-a_{1}-a_{2}-\cdots-a_{k-1}-a_{k}-k-1\right),
\end{array}
$$

Thus, $\quad a_{k+1}-a_{k}=1+\frac{1}{7}\left(-a_{k}-1\right)$, which means $a_{k+1}=\frac{6}{7} a_{k}+\frac{6}{7}$.
Since the fixed point of $f(x)=\frac{6}{7} x+\frac{6}{7}$ is $x=6$, we have $a_{k+1}-6=\frac{6}{7}\left(a_{k}-6\right)$.
Therefore, $a_{k}-6=(a-6)\left(\frac{6}{7}\right)^{k-1}=\frac{1}{7}\left(\frac{6}{7}\right)^{k-1}(m-36)$,
$$
a_{k}=\frac{1}{7}\left(\frac{6}{7}\right)^{k-1}(m-36)+6 .
$$

So $m=a_{1}+a_{2}+\cdots+a_{n}$
$$
\begin{array}{l}
=\frac{1}{7}(m-36)\left[1+\frac{6}{7}+\left(\frac{6}{7}\right)^{2}+\cdots+\left(\frac{6}{7}\right)^{n-1}\right]+6 n \\
=(m-36)\left[1-\left(\frac{6}{7}\right)^{n}\right]+6 n .
\end{array}
$$

Solving for $m$, we get $m=\frac{7^{n}}{6^{n-1}}(n-6)+36$.
Since $7^{n}$ and $6^{n-1}$ are coprime, and $m, n$ are positive integers, it follows that $6^{n-1} \mid (n-6)$. Also, for $n>1$, it is clear that $6^{n-1}>|n-6|$, so the only possibility is $n=6$, hence $m=36$, meaning the sports meet lasted 6 days and a total of 36 medals were issued.