4. Find all three-digit numbers that are equal to the sum of the factorials of their digits!
Solution. We have

$$
\begin{aligned}
& 100a + 10b + c = a! + b! + c! \\
& a! + b! + c! < 1000, \quad 6! = 720 > 99 \text{, so one of } a, b, \text{ or } c \text{ must be 5 or 6.}
\end{aligned}
$$

If one of the numbers is $6$, then $a! + b! + c! > 6! = 720$, so we must exclude 6. Then $\max(a, b, c) = 5$. There are three possibilities:
a) 5 is in the hundreds place: $5bc$, but this is not possible. The largest number obtained is $5! + 5! + 5! = 360 < 500$.

b) 5 is in the tens place: $a5c$

The largest such number is $4! + 5! + 5! = 264$, so the hundreds place must be 1 or 2, i.e.

$$
\begin{array}{ll}
150, & 1! + 5! + 0! \neq 150 \\
151, & 1! + 5! + 1! \neq 151 \\
152, & 1! + 5! + 2! \neq 152 \\
153, & 1! + 5! + 3! \neq 153 \\
154, & 1! + 5! + 4! \neq 154 \\
155, & 1! + 5! + 5! \neq 155 \\
254, & 2! + 5! + 4! < 200 \\
255, & 2! + 5! + 5! \neq 255
\end{array}
$$

c) 5 is in the units place: $ab5$

If it is 1 or 2 in the hundreds place, then similarly as in b), it is only true for $1! + 4! + 5! = 145$.