$33 \cdot 71$ in base $R_{1}$, the fractional expansion of $F_{1}$ is $0.373737 \cdots$, and the fractional expansion of $F_{2}$ is $0.737373 \cdots$. In base $R_{2}$, the expansion of $F_{1}$ is $0.252525 \cdots$, and the expansion of $F_{2}$ is $0.525252 \cdots$. The sum of $R_{1}$ and $R_{2}$ (both expressed in decimal) is
(A) 24.
(B) 22.
(C) 21.
(D) 20.
(E) 19.
(17th American High School Mathematics Examination, 1966)
[Solution] First, write the given fractions as infinite geometric series with common ratios of $\frac{1}{R_{1}}$ and $\frac{1}{R_{2}}$:

And
$$
F_{1}=\frac{3 R_{1}+7}{R_{1}^{2}}+\frac{3 R_{1}+7}{R_{1}^{4}}+\cdots=\frac{3 R_{1}+7}{R_{1}^{2}-1} \text {, }
$$

Therefore,
Similarly,
$$
F_{1}=\frac{2 R_{2}+5}{R_{2}^{2}}+\frac{2 R_{2}+5}{R_{2}^{4}}+\cdots=\frac{2 R_{2}+5}{R_{2}^{2}-1},
$$
$$
\begin{array}{l}
F_{1}=\frac{3 R_{1}+7}{R_{1}^{2}-1}=\frac{2 R_{2}+5}{R_{2}^{2}-1} . \\
F_{2}=\frac{7 R_{1}+3}{R_{1}^{2}-1}=\frac{5 R_{2}+2}{R_{2}^{2}-1} .
\end{array}
$$

Adding the two equations above, we get
$$
F_{1}+F_{2}=\frac{10 R_{1}+10}{\left(R_{1}+1\right)\left(R_{1}-1\right)}=\frac{7 R_{2}+7}{\left(R_{2}+1\right)\left(R_{2}-1\right)} \text {, }
$$

That is
$$
\begin{array}{l}
\frac{10}{R_{1}-1}=\frac{7}{R_{2}-1} . \\
\frac{R_{1}-1}{10}=\frac{R_{2}-1}{7} \\
7 R_{1}-10 R_{2}+3=0 .
\end{array}
$$

Then
That is
Similarly, subtracting the two equations and simplifying, we get
$$
3 R_{1}-4 R_{2}-1=0 .
$$

Solving (1) and (2) gives $R_{1}=11, R_{2}=8$. Therefore, $R_{1}+R_{2}=19$. Hence, the answer is $(E)$.