The number $(2^{48}-1)$ is exactly divisible by two numbers between $60$ and $70$. These numbers are
$\textbf{(A) }61,63\qquad \textbf{(B) }61,65\qquad \textbf{(C) }63,65\qquad \textbf{(D) }63,67\qquad  \textbf{(E) }67,69$
Factor.
\[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)\]
We only care about two terms: $2^{6}+1$ and $(2^{3}+1)(2^{3}-1)$. These simplify to $65$ and $63.$
The answer is $\textbf{(C)}.$
-edited by coolmath34