23. Given that $x$, $y$, and $z$ are positive integers less than 10, and satisfy $(100 x+10 y+z)^{2}=(x+y+z)^{5}$. Then the value of $x^{2}+y^{2}+z^{2}$ is ( ).
(A) 21
(B) 23
(C) 29
(D) 33
(E) 37
23.C.

Let $N=(100 x+10 y+z)^{2}=(x+y+z)^{5}$. Since $N$ is a power of 2 and also a power of 5, it must be a power of 10. Also, because $N$ is the square of a three-digit number and
$$
\begin{array}{l}
2^{10}=\left(2^{5}\right)^{2}=32^{2}, 3^{10}=\left(3^{5}\right)^{2}=243^{2}, \\
4^{10}=\left(4^{5}\right)^{2}=1024^{2},
\end{array}
$$

thus, $N=243^{2}$ (the other numbers are too large or too small).
Therefore, $x=2, y=4, z=3$ is the only solution that satisfies the conditions.