Finding the smallest number $a \geq 0$ such that $a \equiv 7^{83} \bmod 11$.

Question

Find the smallest number $a \geq 0$ such that $a \equiv 7^{83} \bmod 11$.

I know that I am supposed to come to the conclusion $a = 2$, but I don't understand how I reach this conclusion using Fermat's or Euler's theorem.

I have the order $83$ to $7$ in $\mathbb{Z}_{11}$, but I don't know what I write as an answer, because writing just $a = 2$ is not enough to show that I have understood how I came to the answer without using a calculator to just calculate that $7^{83} \bmod 11 = 2$.

What does Fermat's Little Theorem say about $7^{10} \pmod {11}$? — player3236, Oct 30 '20 at 16:03
with numbers of this size even python is enough to get the result — JayTuma, Oct 30 '20 at 16:10

VIVID · Accepted Answer · 2020-10-30T16:15:47.060

2

According to Fermat's Little Theorem $$a^{p-1} \equiv 1 \pmod p$$ for prime $p\nmid a$.

Therefore, $$7^{83} =\color{red}{(7^{11-1})}^8\cdot 7^3 \equiv \color{red}1\cdot 7^3 \equiv 343 \equiv 2\pmod {11}$$ Hence, $a=2$.

edited Oct 30 '20 at 16:15

answered Oct 30 '20 at 16:08

VIVID

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Finding the smallest number $a \geq 0$ such that $a \equiv 7^{83} \bmod 11$.

1 Answers1