5. A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $x!+y!+z!=\overline{x y z}$. Then the value of $x+y+z$ is $\qquad$
5.10.

From $\overline{x y z}=x!+y!+z!$, we get
$$
100 x+10 y+z=x!+y!+z!.
$$

It is easy to see that $x, y, z \leqslant 6$. Otherwise,
$$
x!+y!+z!\geqslant 7!>1000.
$$

If $x=6$, then the left side of equation (1) is $700$, which is a contradiction.
Therefore, $x \leqslant 5$.
Similarly, $y \leqslant 5, z \leqslant 5$.
Thus, $x!+y!+z!\leqslant 3 \times 5!=360$. Hence, $x \leqslant 3$.
When $x=3$,
$$
294+10 y+z=y!+z!\leqslant 2 \times 5!=240,
$$

which is a contradiction.
When $x=2$,
$$
198+10 y+z=y!+z!.
$$

If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 2 \times 4!<198$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,128+z=z$!, which is impossible;
(ii) If $z=5,83+10 y=y$!, then $y \mid 83$, which is also impossible.

When $x=1$, $99+10 y+z=y!+z!$.
If $y \leqslant 4, z \leqslant 4$, then $y!+z!\leqslant 48<99$, which is a contradiction.
Therefore, one of $y, z$ must be 5.
(i) If $y=5,29+z=z$!, which implies $z \mid 29$, which is impossible;
(ii) If $z=5,10 y=y!+16$, by calculation, $y$ can only be 4.

Thus, $x+y+z=10$.