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Difference between revisions of "1972 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
  
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By [[Fermat's little theorem]], we know that <math>2^{100} \equiv 2^{1000 \pmod{12}}\pmod{13}</math>. However, we find that <math>1000 \equiv 4 \pmod{12}</math>, so <math>2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}</math>, so the answer is <math>\boxed{\textbf{(C)}}</math>.
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By Fermat's little theorem, we know that <math>2^{100} \equiv 2^{1000 \pmod{12}}\pmod{13}</math>. However, we find that <math>1000 \equiv 4 \pmod{12}</math>, so <math>2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}</math>, so the answer is <math>\boxed{\textbf{(C)}}</math>.

Revision as of 13:09, 23 June 2021

Problem

When the number $2^{1000}$ is divided by $13$, the remainder in the division is

$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }7\qquad \textbf{(E) }11$

Solution

By Fermat's little theorem, we know that $2^{100} \equiv 2^{1000 \pmod{12}}\pmod{13}$. However, we find that $1000 \equiv 4 \pmod{12}$, so $2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}$, so the answer is $\boxed{\textbf{(C)}}$.

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