0

My instructor has asked to find multiplicative inverse of $27 \pmod{80}$. I am not asking for solution here. But what is meaning of this $27\pmod{80}$? Is it that it is equivalence class in which $27$ element is there? Kindly clarify this. I am new to this modulo stuff.

Thanks

ForgotALot
  • 3,931
Gathdi
  • 1,402
  • $$27\cdot3\equiv1\pmod{80}$$ – lab bhattacharjee Jul 31 '16 at 16:20
  • 1
    By definition, $,27,$ has inverse $,x\iff 27x\equiv 1!\ \pmod{!80}\ , $ To solve it note that $,1\equiv 81,,$ which is divisible by $,27.,,$ Generally you can use the Extended Euclidean Algorithm. – Bill Dubuque Jul 31 '16 at 16:31
  • The notation $27\pmod{80}$ indeed denotes the equivalence class of integers modulo $80$ to which $27$ belongs. – ForgotALot Jul 31 '16 at 17:17
  • There are two mathematical notations in which you can write this. The first one is: $(27\bmod{80})=(107\bmod{80})=(187\bmod{80})=\ldots=((27+80k)\bmod{80})$. The second one is: $27\equiv107\equiv187\equiv\ldots\equiv27+80k\pmod{80}$. – barak manos Jul 31 '16 at 17:35
  • 2
    @labbhattacharjee The OP is explicitly confused by the mod notation and is not asking for a solution. Your terse answer which uses without explanation the same notation is rather unhelpful. – Erick Wong Jul 31 '16 at 18:26
  • @ForgotALot You mean to say that question askes to find inverse of class which leaves remainder 27 when divided by 80 – Gathdi Aug 01 '16 at 11:53

1 Answers1

1

A number $x$ satisfies the congruence $$\ x\equiv a\ \mod\ n\ $$ if and only if $n|x-a$. The inverse , if existent, is the number $y$ with $$\ xy\equiv 1\ \mod\ n$$

This inverse is unique modulo $n$ , but it only exists if $\ gcd(x,n)=1\ $.

Peter
  • 84,454
  • Here it is 27(mod80). So a=27 and n=80. So as far as i understand this i first have to find x and then put in your second equation to find y – Gathdi Aug 01 '16 at 04:10