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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!

Note by Adarsh Kumar
1 year, 8 months ago

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@Satyajit Mohanty @Pi Han Goh Adarsh Kumar · 1 year, 8 months ago

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@Adarsh Kumar The smallest \(n\) is \(17\). Study residue systems modulo \(n\). Satyajit Mohanty · 1 year, 8 months ago

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@Satyajit Mohanty You need to check that 17 is the smallest, though. Otto Bretscher · 1 year, 8 months ago

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@Satyajit Mohanty Kindly tell the method! Adarsh Kumar · 1 year, 8 months ago

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@Adarsh Kumar Oh,wait I got it:did you do it this way:\[10^{18}\equiv1\pmod{19}\\ from\ Euler's\ totient\ function\\ 10^{17}\equiv\dfrac{1}{10}\equiv\dfrac{2}{20}\equiv2\pmod{19}\]? Adarsh Kumar · 1 year, 8 months ago

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@Adarsh Kumar You have only shown that n = 17 is a solution but you didn't show that it's the smallest solution. Pi Han Goh · 1 year, 8 months ago

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@Pi Han Goh How do we do that?Please help! Adarsh Kumar · 1 year, 8 months ago

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@Adarsh Kumar Show that the residues of a prime is (prime - 1). So if you found anything less than a prime, then it's minimum. Pi Han Goh · 1 year, 8 months ago

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@Adarsh Kumar Yup. Satyajit Mohanty · 1 year, 8 months ago

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@Satyajit Mohanty Ok,thanx!Are you in mood for a trigo problem? Adarsh Kumar · 1 year, 8 months ago

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