$10 \cdot 13$ Given $0 \leqslant x_{0}0$, let
$$
x_{n}=\left\{\begin{array}{ll}
2 x_{n-1}, & \text { if } 2 x_{n-1}<1, \\
2 x_{n-1}-1, & \text { if } 2 x_{n-1} \geqslant 1 .
\end{array}\right.
$$

The number of $x_{0}$ such that $x_{0}=x_{5}$ is
(A) 0 .
(B) 1 .
(C) 5 .
(D) 31 .
(E) infinitely many.
(44th American High School Mathematics Examination, 1993)
[Solution] Let $x_{0} \in[0,1)$, in binary form it has the shape
$$
x_{0}=0 . d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots \text { (each } d_{k}=0 \text { or } 1 \text { ). }
$$

From the given recursive relation, we get: moving the binary point one position to the right, and then omitting all the digits before the point. That is,
$$
\begin{array}{c}
x_{0}=0 . d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots \\
x_{1}=0 . d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots .
\end{array}
$$

Therefore, from $x_{0}=x_{5}$, we have
$$
0 . d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots=0 . d_{6} d_{7} d_{8} d_{9} d_{10} d_{11} d_{12} \cdots .
$$

This is true if and only if $x_{0}$ extends with $d_{1} d_{2} d_{3} d_{4} d_{5}$ as a repeating cycle, and there are $2^{5}=32$ such cycles.

However, when $d_{1}=d_{2}=d_{3}=d_{4}=d_{5}=1$, we have $x_{0}=1$. Therefore, for $x_{0} \in [0,1)$ that satisfies $x_{0}=x_{5}$, there are $32-1=31$ values.
Hence, the answer is $(D)$.