43. Find all three-digit numbers that are equal to the arithmetic mean of all numbers obtained from the given number by all possible permutations of its digits (including, of course, the "identity permutation" that leaves all the digits of the number in place).
43. Let \( N = \overline{xyz} = 100x + 10y + z \) be the desired number, where \( x, y, z \) are the digits of the number. By permuting the digits of \( N \), we obtain the numbers \( N_1 = \overline{yxz} = 100y + 10x + z, \ldots, N_5 = \overline{zyx} = 100z + 10y + x \). The sum \( N + N_1 + \ldots + N_5 \) of all these numbers should equal six times the number \( N \), from which we easily obtain:

\[
(2 \cdot 100 + 2 \cdot 10 + 2)(x + y + z) = 6(100x + 10y + z)
\]

(because in the six numbers \( N, N_1, N_2, N_3, N_4, N_5 \), each digit, for example, \( x \), appears 2 times in the first position, 2 times in the second position, and 2 times in the third position), or

\[
37(x + y + z) = 100x + 10y + z
\]

i.e.,

\[
63x = 27y + 36z
\]

Dividing both sides by 9, we get:

\[
7x = 3y + 4z, \text{ or } 7(x - y) = 4(z - y).
\]

But the differences \( x - y \) and \( z - y \) in absolute value do not exceed 9, from which it follows that the last equality is possible only if

\( x - y = 0, z - y = 0 \), or \( x - y = 4, z - y = 7 \), or

\[
x - y = -4, z - y = -7
\]

If \( z - y = 7 \), then \( z = 9, 8 \) or 7; if \( z - y = -7 \), then \( y = 9, 8 \) or 7; therefore (discarding the dubious "solution" \( N = 000 \)), we obtain 15 possible values of the number \( N \):

\( N = 111, 222, 333, 444, 555, 666, 777, 888, 999, 407 \),

\( 518, 629, 370, 481 \) or 592.