(7) A three-digit number $\overline{x y z}, 1 \leqslant x \leqslant 9,0 \leqslant y, z \leqslant 9$, and $\overline{x y z}=x!+$ $x!+z!$. Then the value of $x+y+z$ is $\qquad$ .
(7) 10 Hint: Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$. And $1!=1$, $2!=2$, $3!=6$, $4!=24$, $5!=120$. Therefore, $x=1, y=4, z=5$, so $x+y+z=10$.