Hello,

Using a calculator,I know that the solution to $\displaystyle d = 55^{-1} \ (mod \ 10752) $ is d = 391.

Can someone please show me how to prove this by hand using Euclid's Algorithm? Any help would be greatly appreciated.

Thanks

Printable View

- Mar 30th 2009, 04:56 AMJimmy_WHow can I work this out?
Hello,

Using a calculator,I know that the solution to $\displaystyle d = 55^{-1} \ (mod \ 10752) $ is d = 391.

Can someone please show me how to prove this by hand using Euclid's Algorithm? Any help would be greatly appreciated.

Thanks - Mar 30th 2009, 05:59 AMCaptainBlack
- Mar 30th 2009, 07:08 AMHallsofIvy
Captain Black, he said that he knew that "using a calculator" and asked how you would find it "by hand".

Jimmy W, if $\displaystyle d= 55^{-1}$ (mod 10752) then 55d= 10752k+ 1 for some integer k. That is the same as 55d- 10752k= 1 so, using Euclid's algorithm:

55 divides into 10752 195 times with remainder 27. 27 divides into 55 twice with remainder 1. That is: 55 -2(27)= 1 and, since 10752- 195(55), that is 55- 2(10752- 195(55))= (1+ 2(195))(55)- 2(10752)= 391(55)- 2(10752)= 1. That means that one solution is d= 391, k= 2.