Difference between revisions of "1972 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
− | 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>. | + | 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 14:09, 23 June 2021
Problem
When the number is divided by
, the remainder in the division is
Solution
By Fermat's little theorem, we know that . However, we find that
, so
, so the answer is
.