Place $n$ balls numbered $1,2, \cdots, n(n \geqslant 6)$ in sequence on a circle. Starting from the 1st ball, remove 1 ball every 5 balls, i.e., remove the balls numbered $6,12, \cdots, 6\left[\frac{n}{6}\right], 6\left(\left[\frac{n}{6}\right]+1\right)-n$, $\cdots$ (where $[x]$ represents the integer part of $x$), until 5 balls are left. It is known that when $n=56$, the numbers of the 5 remaining balls are $10,20,29,37,56$. Find the numbers of the 5 remaining balls when $n=57, n=58, n=59, n=60$.
Solution: Place $n$ balls numbered $1, 2, \cdots, n$ in a circular arrangement, and remove the balls according to the specified method. Let the original numbers of the 5 remaining balls be $F_{i}(n), i=1,2,3,4,5$.
First, we find the relationship between $F_{1}(n+1)$ and $F_{i}(n)$.
Place $n+1$ balls numbered $1, 2, \cdots, n+1$ in a circular arrangement, and remove the 6th ball. Re-number the remaining $n$ balls as follows: the original 7th ball is the new 1st, the original 8th ball is the new 2nd, $\cdots$, the original $(n+1)$-th ball is the new $(n-5)$-th, the original 1st ball is the new $(n-4)$-th, $\cdots$, the original 5th ball is the new $n$-th. Thus, we obtain a transformation $\sigma$ from the old numbers to the new numbers of these $n$ balls:
$$
\sigma(i)=\left\{\begin{array}{ll}
i-6, & 7 \leqslant i \leqslant n+1, \\
i+n-5, & 1 \leqslant i \leqslant 5 .
\end{array}\right.
$$

Now consider the $n$ re-numbered balls remaining on the circle.
Remove the balls according to the specified method, and the new numbers of the 5 remaining balls are $F_{1}(n), i=1,2,3,4,5$. Note that these 5 balls are exactly the balls that remain when the original $n+1$ balls are numbered and the balls are removed according to the specified method. Therefore, the numbers of the 5 remaining balls when the original $n+1$ balls are numbered and the balls are removed according to the specified method, $F_{1}(n+1)$, are the pre-images of $F_{i}(n)$ under $\sigma$, $i=1,2,3,4,5$.
From equation (1), it is easy to obtain the inverse transformation of $\sigma$:
$$
\sigma^{-1}(i)=\left\{\begin{array}{ll}
i+6, & 1 \leqslant i \leqslant n-5, \\
i-(n-5), & n-4 \leqslant i \leqslant n .
\end{array}\right.
$$

Thus, the required recurrence relation is
$$
F_{i}(n+1)=\left\{\begin{array}{l}
F_{i}(n)+6, \quad 1 \leqslant F_{i}(n) \leqslant n-5, \\
F_{1}(n)+5-n, \quad n-4 \leqslant F_{i}(n) \leqslant n .
\end{array}\right.
$$

Using this recurrence relation, it is easy to calculate: when $n=57$, the numbers of the 5 remaining balls are $5, 16, 26, 35, 43$; when $n=58$, the numbers of the 5 remaining balls are $11, 22, 32, 41, 49$; when $n=59$, the numbers of the 5 remaining balls are $17, 28, 38, 47, 55$; when $n=60$, the numbers of the 5 remaining balls are $1, 23, 34, 44, 53$.
(Fang Tinggang, Chengdu No. 7 High School, 610041)