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

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== Solution ==
 
== Solution ==
Factor.
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Factor 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^{3}+1)(2^{3}-1)</cmath>
  
We only care about two terms: <math>2^{6}+1</math> and <math>(2^{3}+1)(2^{3}-1)</math>. These simplify to <math>65</math> and <math>63.</math>
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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>.
  
The answer is <math>\textbf{(C)}.</math>
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Thus, our answer is <math>\boxed{\textbf{(C) }63,65}.</math>
  
 
-edited by coolmath34
 
-edited by coolmath34
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== See Also ==
 +
{{AHSME 35p box|year=1971|num-b=13|num-a=15}}
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{{MAA Notice}}

Revision as of 11:53, 1 August 2024

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 with difference of squares. \[2^{48}-1 = (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)\]

We only care about two terms: $2^{6}+1$ and $(2^6-1)$. 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
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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