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Find the inverse of $2^{10}$ modulo $2011$ and then compute the remainder of the division of $2^{2000}$ by $2011$.

I managed to find the inverse, it was simple, I just had to solve the diophantine equation $1024x+2011k = 1$. I did this by applying the extended euclidean algorithm, and found that the inverse $s$ is $s=83$. But I'm not sure how to apply it to solve $2^{2000}\pmod{2011}$. Any help is appreciated.

Bill Dubuque
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A. Riba
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  • By Fermat $,2^{\color{#c00}{2010}}\equiv 1,$ so $\bmod \color{#c00}{2010}!:\ \color{#0a0}{2000\equiv -10},\Rightarrow, 2^{\color{#0a0}{2000}}\equiv 2^{\color{#0a0}{-10}},$ by mod order reduction in linked dupe. – Bill Dubuque Jul 03 '23 at 01:40

1 Answers1

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ok, you know that $$2^{\varphi(2011)}\equiv 1\; \bmod 2011$$ but $$\varphi(2011)=2010=10+2000$$ then you can use this, and obtain $$2^{10}*2^{2000} = 1 \;\bmod 2011$$ then you have the answer

J. W. Tanner
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Pablo Maurente
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