Find $x$ such that $x^{677} ≡ 3 \mod 2020$.

Question

I have been reviewing number theory questions and there is one problem I am stuck on.

My approach was to start by applying Euler’s Theorem. I know that $\phi(2020) = 800$, but I don’t know if this is very useful... How should I proceed?

Since $2020=2^2\times 5\times 101$ you can solve the problem mod $4,5,101$ separately and then use the Chinese Remainder Theorem. — lulu, May 22 '20 at 23:05
"The answer is x≡1(mod2020) if that helps" As $1^{677}\not \equiv 3 \pmod{2020}$ that obviously isn't correct. — fleablood, May 22 '20 at 23:14
@AlexeyBurdin: we want $x^{6\color{red}77}\equiv\color{red}3$, not $x^{667}\equiv1$ — J. W. Tanner, May 22 '20 at 23:15

score 0 · Answer 1 · answered May 22 '20 at 23:13

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As suggested in the comment by lulu, $2020=2^2\times5\times101$,

so solve the problem mod $4,5,$ and $101$ separately,

and then use the Chinese remainder theorem.

I.e., $x\equiv3\bmod4$ , $x\equiv3\bmod 5$, and $x^{77}\equiv3\bmod101$.

$77\times13=1001\equiv1\bmod100$, so $x\equiv3^{13}\equiv38\bmod101$.

So we have $x\equiv3\bmod20$ and $x\equiv38\bmod101$.

Can you now show $x\equiv543\bmod2020$?

answered May 22 '20 at 23:13

J. W. Tanner

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If it were $x^{6\color{red}67}\equiv3$, then we'd have $x\equiv3\bmod4, x\equiv3^3\bmod5, $ and $x\equiv3^3\bmod101$, so some arithmetic would be easier – J. W. Tanner May 24 '20 at 02:06

Find $x$ such that $x^{677} ≡ 3 \mod 2020$.

1 Answers1