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!

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

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Satyajit Mohanty @Pi Han Goh

Log in to reply

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

Log in to reply

Kindly tell the method!

Log in to reply

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

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply