$33 \cdot 15$ For $n>1, x_{n}=\frac{2^{n}}{x_{n-1}}$. Then the last digit of the product $x_{1} \cdot x_{2} \cdots \cdots x_{200}$ is
(A) 2.
(B) 4.
(C) 6.
(D) 8.
(8th "Jinyun Cup" Junior High School Mathematics Invitational Competition, 1991)
[Solution] Given $x_{n} \cdot x_{n-1}=2^{n}$, or $x_{n-1} x_{n}=2^{n}$. Then
$$
\begin{aligned}
x_{1} \cdot x_{2} \cdot x_{3} \cdot \cdots \cdot x_{200} & =\left(x_{1} x_{2}\right)\left(x_{3} x_{4}\right) \cdots\left(x_{199} x_{200}\right) \\
& =2^{2} \cdot 2^{4} \cdots \cdots \cdot 2^{200} \\
& =2^{2+4+\cdots+200}=2^{10100} \\
& =4^{5050},
\end{aligned}
$$
Since the last digit of powers of 4 changes periodically as $4,6,4,6, \cdots$.
Therefore, the last digit (units digit) of $4^{5050}$ is 6.
Hence, the correct choice is $(C)$.