1971 AHSME Problems/Problem 14

Problem

The number $(2^{48}-1)$ is exactly divisible by two numbers between $60$ and $70$. These numbers are

$\textbf{(A) }61,63\qquad \textbf{(B) }61,65\qquad \textbf{(C) }63,65\qquad \textbf{(D) }63,67\qquad  \textbf{(E) }67,69$

Solution

Factor repeatedly with difference of squares. \[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^6-1)\]

We only care about two terms: $2^6+1$ and $2^6-1$. These simplify to $65$ and $63$.

Thus, our answer is $\boxed{\textbf{(C) }63,65}.$

-edited by coolmath34

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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