1. Determine all natural numbers less than 1000 that are equal to the sum of the factorials of their digits.
Solution. We need to determine all triples $(x, y, z)$ for which

\[
\begin{aligned}
& x, y, z \in\{0,1,2, \ldots, 9\},(x, y, z) \neq(0,0,0) \\
& 100 x+10 y+z=x!+y!+z!
\end{aligned}
\]

If $(x, y, z)$ is such a triple, then $x \leq 5, y \leq 5, z \leq 5$ and $x!+y!+z! \leq 3 \cdot 5! = 360$. Indeed, otherwise, $x!+y!+z! \geq 6! = 720$, and if $x! \geq 7! = 5040$, which is a contradiction. By checking the numbers 55, 155, 255, 355, 455, it is easily seen that none of them are solutions to the problem, which means that at most one of the numbers $x, y, z$ is equal to 5.

a) Suppose exactly one of the numbers $x, y, z$ is equal to 5. It is easily verified that the other two numbers are not equal to 4. Therefore,

\[
120 \leq x!+y!+z! \leq 5! + 4! + 3! = 150
\]

By checking the numbers $125, 135, 145$, we find that 145 is a solution to the problem.

b) Suppose $x \leq 4, y \leq 4, z \leq 4$. By checking, it is found that in this case the triples $(0,0,1)$ and $(0,0,2)$ are the only solutions to the problem.

Finally, the required numbers are 1, 2, and 145.