$1 \cdot 70$ radical $\sqrt{\frac{8^{10}+4^{10}}{8^{4}+4^{11}}}$ is equal to
(A) $\sqrt{2}$.
(B) 16 .
(C) 32 .
(D) 12 .
(E) 512.5 .
(44th American High School Mathematics Examination, 1993)
[Solution]
$$
\begin{aligned}
\sqrt{\frac{8^{10}+4^{10}}{8^{4}+4^{11}}} & =\sqrt{\frac{\left(2^{3}\right)^{10}+\left(2^{2}\right)^{10}}{\left(2^{3}\right)^{4}+\left(2^{2}\right)^{11}}}=\sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}} \\
& =\sqrt{\frac{2^{8}\left(2^{12}+2^{22}\right)}{2^{12}+2^{22}}}=\sqrt{2^{8}}=16 .
\end{aligned}
$$

Therefore, the answer is (B)