$10 \cdot 89$ When $a_{1}, a_{2}, \cdots, a_{25}$ are 0 or 2, all numbers $x$ of the form $x=\frac{a_{1}}{3}+\frac{a_{2}}{3^{2}}+\cdots+\frac{a_{22}}{3^{25}}$ can satisfy
(A) $0 \leqslant x<\frac{1}{3}$.
(B) $\frac{1}{3} \leqslant x<\frac{2}{3}$.
(C) $\frac{2}{3} \leqslant x<1$.
(D) $0 \leqslant x<\frac{1}{3}$ or $\frac{2}{3} \leqslant x<1$.
(E) $\frac{1}{2} \leqslant x \leqslant \frac{3}{4}$.
(25th American High School Mathematics Examination, 1974)
[Solution] By the formula for the sum of an infinite geometric series, we have
$$
0 \leqslant x<\sum_{n=1}^{\infty} \frac{2}{3^{n}}=\frac{\frac{2}{3}}{1-\frac{1}{3}}=1 \text {. }
$$

If $a_{1}=0$, then $0 \leqslant x<\sum_{n=2}^{\infty} \frac{2}{3^{n}}=\frac{\frac{2}{9}}{1-\frac{1}{3}}=\frac{1}{3}$; if $a_{1}=2$, then $\frac{2}{3}+0+0+\cdots \leqslant x<1$. From the above, we have $0 \leqslant x<\frac{1}{3}$ or $\frac{2}{3} \leqslant x<1$. Therefore, the answer is $(D)$.