My doubt has been resolved by Satyajit Mohanty and Pi Han Goh.$$How would you go about solving this problem?:$10^n \equiv 2\pmod{19};n \in \mathbb{Z}^{+}\\
Smallest\ such\ n=?$Please help!

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@Adarsh Kumar
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Oh,wait I got it:did you do it this way:$10^{18}\equiv1\pmod{19}\\
from\ Euler's\ totient\ function\\
10^{17}\equiv\dfrac{1}{10}\equiv\dfrac{2}{20}\equiv2\pmod{19}$?

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TopNewest@Satyajit Mohanty @Pi Han Goh

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The smallest $n$ is $17$. Study residue systems modulo $n$.

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Kindly tell the method!

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$10^{18}\equiv1\pmod{19}\\ from\ Euler's\ totient\ function\\ 10^{17}\equiv\dfrac{1}{10}\equiv\dfrac{2}{20}\equiv2\pmod{19}$?

Oh,wait I got it:did you do it this way:Log in to reply

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You need to check that 17 is the smallest, though.

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