1. 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
$$abc + \overline{acb} + \overline{bac} + \overline{bca} + \overline{cab} + \overline{cba} = 6 \overline{abc}$$,
which simplifies to $7a = 3b + 4c$.
$$7(a-b) = 4(c-b)$$

From this, either $a = b = c$; or $a - b = 4, c - b = 7$ (then $0 \leqslant b \leqslant 2$); or $b - a = 4, b - c = 7$ (then $7 \leqslant b \leqslant 9$).

Therefore, there are 15 numbers that satisfy the condition, which are
$$111, 222, \cdots, 999, 407, 518, 629, 370, 481, 592$$