Find all positive integer solutions to the equation $w! = x! + y! + z!$ . (1983 Canadian Mathematical Olympiad problem)
Let's assume $x \leqslant y \leqslant z$, obviously $w \geqslant z+1$.
Then we have
$$
(z+1)!\leqslant w!=x!+y!+z!\leqslant 3 z!,
$$

which implies $z+1 \leqslant 3$, or $z \leqslant 2$. At this point, $x, y$ are both no greater than 2.
We can consider the following four cases:
$$
\begin{array}{l}
z=1, x=1, y=1, \\
z=2, x=1, y=1, \\
z=2, x=1, y=2, \\
z=2, x=2, y=2.
\end{array}
$$

The only solution is $x=2, y=2, z=2, n=3$.