$29 \cdot 26\left(2^{48}-1\right)$ can be divided by two numbers between 60 and 70, these two numbers are
(A) 61,63 .
(B) 61,65 .
(C) 63,65 .
(D) 63,67 .
(E) 67,69 .
(22nd American High School Mathematics Examination, 1971)
[Solution] Notice the transformation of the following algebraic expression:
$$
\begin{aligned}
2^{48}-1 & =\left(2^{24}-1\right)\left(2^{24}+1\right) \\
& =\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right) \\
& =\left(2^{6}-1\right)\left(2^{6}+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right) \\
& =63 \cdot 65 \cdot\left(2^{12}+1\right)\left(2^{24}+1\right) .
\end{aligned}
$$

Therefore, the answer is $(C)$.