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I am trying to find $10^{1001}(\text{mod} \ 1001)$. It seemed appropriate to use Euler's theorem, as $\text{gcd}(10,1001)=1$. I have found that:

\begin{align} \phi(1001)&=\phi(7)\phi(11)\phi(13)=720 \ \ \ (\text{where $\phi$} \ \text{denotes Euler's phi-function}) \\ \implies 10^{1001}(\text{mod} \ 1001)&=10^{720+281} (\text{mod} \ 1001) \\ &=1^{720}\cdot10^{281}(\text{mod} \ 1001) \\ &=10^{281}(\text{mod} \ 1001). \end{align}

I don't know how to solve the problem from here. Any advice would be appreciated.

Bill Dubuque
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M B
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1 Answers1

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Hint: $10^3\equiv-1\bmod 1001,$ so $10^{3n+2}\equiv(-1)^n100\bmod 1001$

J. W. Tanner
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