4-192 Find such a three-digit number, it equals the sum of the factorials of its digits.



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[Solution] Let $M=\overline{abc}=100a+10b+c$ be the desired three-digit number. According to the problem, we have $\overline{abc}=a!+b!+c!$

Analyzing both sides of (1), considering
$$
4!=24, 5!=120, 6!=720, 7!=5040,
$$

we can conclude that none of the three digits of $M$ exceed 6. Therefore, $M$ does not exceed 666. This implies that each digit of $M$ does not exceed 5. From equation (1), the number $a!+b!+c!$ must be a three-digit number, so at least one of $a, b, c$ must equal 5, otherwise $a!+b!+c! \leqslant 4!+4!+4!=72$ is less than any three-digit number.
From the derived range of $a, b, c$, we get
$$
a!+b!+c! \leqslant 5!+5!+5!=360.
$$

Thus, from equation (1), the desired number $M=\overline{abc} \leqslant 360$. Therefore, $a \leqslant 3$.
It is not difficult to prove that $a=1$. Indeed, when $a=3$, the number on the left side of equation (1) is greater than 300 ($\overline{3bc}>300$) because one of $b$ or $c$ equals 5, while the number on the right side does not exceed 3! + 5! + 5! = 246, which is less than 300. Therefore, equation (1) does not hold when $a=3$.

If $a=2$, then for equation (1) to hold, it must be that $b=c=5$, because in the opposite case, the number on the right side of equation (1)
$$
a!+b!+c! \leqslant 2!+5!+4!=146 < \overline{1bc}=M.
$$

The digit $b$ cannot equal 5, because in the opposite case, we would have
$$
a!+b!+c! \leqslant 1!+5!+4!=145 < \overline{15c}=M.
$$

Thus, $a=1$ and $c=5$. To determine the digit $b$, equation (1) becomes $\overline{1b5}=1!+b!+5!$ or $\overline{1b5}=2+b!$
For this equation to hold, the last digit of the number $2+b!$ must be 5, which is only true when $b=4$.
Therefore, the desired number $M=145$.