Reassurance with my math

Finally found myself just playing around with my method for finding a modular inverse and was reassuring. Math is that way. Doing a post to let know am still active and to show it again.

So will calculate modular inverse of 19 mod 1001 which is 7(11)(13).

Let 2m = 1 mod 1001, then F₀ = 19(19 + y₀) mod 1001. Can pick y₀ = 34. Then F₀ = 6 mod 1001. That was lucky! Can pick whatever I like. 

Picking so that F₀ is as small as possible without 19 as a factor but coprime to 1001, and usually regardless have found need to recurse with coefficient of d as new modulus.

6d = (6n - 7)(53) mod 1001

6d = 318n - 371 mod 1001 

And 6(167) = 1002 = 1 mod 1001 which is very convenient.

So: d = 53n - 896 mod 1001, so can use n = 1 and then d = -843. So 19^-1 = 843 mod 1001.

Those values worked better than I expected, as of course really like easy. But yeah is very reassuring to just do some math and get the correct answer. 

My way makes calculating the modular inverse so straightforward.

Am still active on this blog but often just reviewing. Have a lot of research results! Can play around endlessly with them.