Three. (50 points) Place $n$ balls numbered $1,2, \cdots, n(n \geqslant 6)$ in sequence on a circle. Starting from the 1st number, remove one 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 $[x]$ denotes the integer part of $x$), until only 5 balls remain. It is known that when $n=56$, the numbers of the remaining 5 balls are $10,20,29,37,56$. Find the numbers of the remaining 5 balls when $n=57,58,59,60$.
Three, Solution: Place $n$ balls numbered $1,2, \cdots, n(n \geqslant 6)$ in sequence on a circle, and remove the balls according to the rules specified in the problem. 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 $F_{i}(n+1)$ and $F_{i}(n)$.

Place $n+1$ balls numbered $1,2, \cdots, n+1$ in sequence on a circle, and remove the 6th ball. Re-number the remaining $n$ balls as follows: the original 7th ball becomes the new 1st, the original 8th ball becomes the new 2nd, $\cdots$, the original $(n+1)$th ball becomes the new $(n-5)$th, the original 1st ball becomes the new $(n-4)$th, $\cdots$, the original 5th ball becomes the new $n$th. Thus, we obtain a transformation of the old numbers to the new numbers: $\sigma(i)=$ $\left\{\begin{array}{cc}i-6, & 7 \leqslant i \leqslant n+1 \\ i+n-5, & 1 \leqslant i \leqslant 5\end{array}\right.$.
Now consider the $n$ re-numbered balls remaining on the circle.
Remove the balls according to the rules specified in the problem, and the new numbers of the 5 remaining balls are $F_{i}(n), i=1,2,3,4,5$. 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 rules specified in the problem. Therefore, the numbers of the 5 remaining balls when the original $n+1$ balls are numbered and the balls are removed according to the rules specified in the problem, $F_{i}(n+1)$, are exactly the pre-images of $F_{i}(n)$ under $\sigma$, $i=1,2,3,4,5$.
It is easy to obtain the inverse transformation of $\sigma$ as $\sigma^{-1}(i)=\left\{\begin{array}{cc}i+6, & 1 \leqslant i \leqslant n-5 \\ i-n+5, & n-4 \leqslant i \leqslant n\end{array}\right.$.
Thus, the recursive relation we seek is $F_{i}(n+1)=\left\{\begin{array}{lc}F_{i}(n)+6, & 1 \leqslant F_{i}(n) \leqslant n-5 \\ F_{i}(n)-n+5, & n-4 \leqslant F_{i}(n) \leqslant n\end{array}\right.$.
Using this recursive 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 $11,23,34,44,53$.