Question 22 Given $0 \leqslant x_{0}0$, let
$$
x_{n}=\left\{\begin{array}{ll}
2 x_{n-1} & \text { if } 2 x_{n-1}<1 ; \\
2 x_{n-1}-1 & \text { if } 2 x_{n-1} \geqslant 1,
\end{array}\right.
$$

then the number of $x_{0}$ for which $x_{0}=x_{5}$ is:
(A) 0 ;
(B) 1 ;
(C) 5 ;
(D) 31 ;
(E) infinitely many.
(1993 44th American High School Mathematics Examination)
Solve for $x_{0} \in[0,1]$, when represented in binary form as
$$
\begin{array}{c}
x_{0}=0 . d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots, \\
x_{1}=0 . d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots .
\end{array}
$$

Then, by $x_{0}=x_{5}$, we have
$$
\text { 0. } d_{1} d_{2} d_{3} d_{4} d_{5} d_{6} d_{7} \cdots=0 . d_{6} d_{7} d_{8} d_{9} \cdots \text {. }
$$

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, giving a total of $2^{5}=32$ 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 satisfy $x_{0}=x_{5}$, there are $32-1=31$ values. Hence, the answer is (D).