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
5 years, 8 months ago

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Comments

@Satyajit Mohanty @Pi Han Goh

Adarsh Kumar - 5 years, 8 months ago

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The smallest nn is 1717. Study residue systems modulo nn.

Satyajit Mohanty - 5 years, 8 months ago

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Kindly tell the method!

Adarsh Kumar - 5 years, 8 months ago

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@Adarsh Kumar Oh,wait I got it:did you do it this way:10181(mod19)from Eulers totient function10171102202(mod19)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 - 5 years, 8 months ago

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@Adarsh Kumar Yup.

Satyajit Mohanty - 5 years, 8 months ago

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@Satyajit Mohanty Ok,thanx!Are you in mood for a trigo problem?

Adarsh Kumar - 5 years, 8 months ago

@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 - 5 years, 8 months ago

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@Pi Han Goh How do we do that?Please help!

Adarsh Kumar - 5 years, 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 - 5 years, 8 months ago

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

Otto Bretscher - 5 years, 8 months ago
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