II. (50 points) An old man divides his savings of $m$ gold coins among his $n$ children ($m, n$ are positive integers greater than 1). First, he gives the eldest child 1 gold coin and $\frac{1}{7}$ of the remainder; then, from the remaining coins, he gives the second child 2 gold coins and $\frac{1}{7}$ of the remainder; and so on, the $n$-th child receives $n$ gold coins and $\frac{1}{7}$ of the remainder, until the youngest child receives the last $n$ gold coins. Does the old man give each child the same number of gold coins?
Let the number of gold coins left after giving to the $k$-th child be $a_{k}$, then $a_{0}=m, a_{n-1}=n$,
$$
a_{k}=a_{k-1}-\left(k+\frac{a_{k-1}-k}{7}\right)=\frac{6}{7}\left(a_{k-1}-k\right) \text {. }
$$

Thus, $a_{k}+6 k-36=\frac{6}{7}\left[a_{k-1}+6(k-1)-36\right]$.
This indicates that the sequence $b_{k}=a_{k}+6 k-36$ is a geometric sequence with common ratio $q=\frac{6}{7}$, where the first term and the last term are
$b_{0}=a_{0}-36=m-36$,
$b_{n-1}=a_{n-1}+6(n-1)-36=7 n-42$.
Substituting into the general term formula $b_{n-1}=b_{0} q^{n-1}$ gives
$7 n-42=(m-36)\left(\frac{6}{7}\right)^{n-1}$,

which means $m=36+\frac{7^{n}(n-6)}{6^{n-1}}$.
Since $m$ is a positive integer, $6^{n-1} \mid 7^{n}(n-6)$.
Since $7^{n}$ and $6^{n-1}$ are coprime, it follows that $6^{n-1} \mid(n-6)$.
But $-6^{n-1}+6 \leqslant 01)$.
Solving this yields $n=6$. Thus, $m=36$.
Therefore,
the eldest child gets $1+\frac{36-1}{7}=6$ coins,
the second child gets $2+\frac{30-2}{7}=6$ coins,
the third child gets $3+\frac{24-3}{7}=6$ coins,
the fourth child gets $4+\frac{18-4}{7}=6$ coins,
the fifth child gets $5+\frac{12-5}{7}=6$ coins,
and the youngest child gets the remaining 6 coins.
So, the old man gives each child the same number of gold coins.
Note: This problem provides a prototype for the 6th question of the 9th IMO.