1971 AHSME Problems/Problem 14

Problem

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

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

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

1971 AHSC (Problems • Answer Key • Resources)
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