Waste less time on Facebook — follow Brilliant.
×

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
2 years, 1 month ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

@Satyajit Mohanty @Pi Han Goh

Adarsh Kumar - 2 years, 1 month ago

Log in to reply

The smallest \(n\) is \(17\). Study residue systems modulo \(n\).

Satyajit Mohanty - 2 years, 1 month ago

Log in to reply

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

Otto Bretscher - 2 years, 1 month ago

Log in to reply

Kindly tell the method!

Adarsh Kumar - 2 years, 1 month ago

Log in to reply

@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 - 2 years, 1 month ago

Log in to reply

@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 - 2 years, 1 month ago

Log in to reply

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

Adarsh Kumar - 2 years, 1 month ago

Log in to reply

@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 - 2 years, 1 month ago

Log in to reply

@Adarsh Kumar Yup.

Satyajit Mohanty - 2 years, 1 month ago

Log in to reply

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

Adarsh Kumar - 2 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...