10.59 Find all such three-digit numbers that are equal to the sum of the factorials of their digits.
(6th Moscow Mathematical Olympiad, 1940)
[Solution] Let $\overline{a b c}=a!+b!+c!$.
Since $7!=5040$ is a four-digit number, there are no numbers greater than 6 in $a, b, c$.
Moreover, if one of $a, b, c$ is 6, then because $6!=720$, this would result in $a b c>6!$ $=720$, and at least one of $a, b, c$ would be greater than 6, which is impossible.
Additionally, it is not possible that
$$
555=5!+5!+5!,
$$

so $a, b, c$ cannot all be 5.
Also, $4!+4!+4!<100$ is not a three-digit number, so at least one of $a, b, c$ must be 5.
By $3 \cdot 5!=360$, then $a \leqslant 3$.
If $a=1$, then we have
$$
145=1!+4!+5!.
$$

If $a=2$, then $b=c=5$ must hold, in which case
$$
255 \neq 2!+5!+5!.
$$

If $a=3$, then $b=c=5$ must hold, in which case
$$
355 \neq 3!+5!+5!
$$

Therefore, there is only one three-digit number with the given property, which is
$$
145=1!+4!+5!
$$