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# Help needed in NT!

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!

1 year, 1 month ago

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@Satyajit Mohanty @Pi Han Goh · 1 year, 1 month ago

The smallest $$n$$ is $$17$$. Study residue systems modulo $$n$$. · 1 year, 1 month ago

You need to check that 17 is the smallest, though. · 1 year, 1 month ago

Kindly tell the method! · 1 year, 1 month ago

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

You have only shown that n = 17 is a solution but you didn't show that it's the smallest solution. · 1 year, 1 month ago

Show that the residues of a prime is (prime - 1). So if you found anything less than a prime, then it's minimum. · 1 year, 1 month ago

Yup. · 1 year, 1 month ago