Difference between revisions of "1971 AHSME Problems/Problem 14"

Latest revision as of 11:54, 1 August 2024

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)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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 == Solution ==
-Factor with [[difference of squares]].
+Factor repeatedly with [[difference of squares]].
-<cmath>2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)</cmath>
+<cmath>2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^6-1)</cmath>
 We only care about two terms: <math>2^6+1</math> and <math>2^6-1</math>. These simplify to <math>65</math> and <math>63</math>.