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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TopNewest@Satyajit Mohanty @Pi Han Goh – Adarsh Kumar · 1 year, 4 months ago

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– Satyajit Mohanty · 1 year, 4 months ago

The smallest \(n\) is \(17\). Study residue systems modulo \(n\).Log in to reply

– Otto Bretscher · 1 year, 4 months ago

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

– Adarsh Kumar · 1 year, 4 months ago

Kindly tell the method!Log in to reply

– Adarsh Kumar · 1 year, 4 months ago

Oh,wait I got it:did you do it this way:\[10^{18}\equiv1\pmod{19}\\ from\ Euler's\ totient\ function\\ 10^{17}\equiv\dfrac{1}{10}\equiv\dfrac{2}{20}\equiv2\pmod{19}\]?Log in to reply

– Pi Han Goh · 1 year, 4 months ago

You have only shown that n = 17 is a solution but you didn't show that it's the smallest solution.Log in to reply

– Adarsh Kumar · 1 year, 4 months ago

How do we do that?Please help!Log in to reply

– Pi Han Goh · 1 year, 4 months ago

Show that the residues of a prime is (prime - 1). So if you found anything less than a prime, then it's minimum.Log in to reply

– Satyajit Mohanty · 1 year, 4 months ago

Yup.Log in to reply

– Adarsh Kumar · 1 year, 4 months ago

Ok,thanx!Are you in mood for a trigo problem?Log in to reply