$1 \cdot 178$ Find all three-digit numbers $A$ such that the arithmetic mean of all numbers obtained by rearranging the digits of $A$ is still equal to $A$.


[Solution] Let $\overline{abc}$ be the number we are looking for. Then, according to the problem, we have
$$
\overline{abc}+\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=6 \overline{abc} \text {, }
$$

which simplifies to $222(a+b+c)=6(100a+10b+c)$,
and further simplifies to $\quad 7a=3b+4c$.
$$
7(a-b)=4(c-b) .
$$

From this, either $a=b=c$; or $a-b=4, c-b=7$ (in which case $0 \leqslant b \leqslant 2$); or $b-a=4, b-c=7$ (in which case $7 \leqslant b \leqslant 9$).
Therefore, there are 15 numbers that satisfy the condition, which are
$$
111,222, \cdots, 999,407,518,629,370,481,592 .
$$