# mod arithmetic

• May 14th 2008, 02:36 AM
hunkydory19
mod arithmetic
Consider the affine enciphering transformation:

f(x) = 9x + 4 mod 49

on single letter message blocks of the ordinary English alphabet A - Z
and the letter b for “space”.

How many more letters can we add to our alphabet without any
further changes to our encoding procedure?

Would this just be 26 x 2 = 52
52 - 49 = 3, hence 3 more letters could be added?
• May 14th 2008, 10:32 AM
Isomorphism
Quote:

Originally Posted by hunkydory19
Consider the affine enciphering transformation:

f(x) = 9x + 4 mod 49

on single letter message blocks of the ordinary English alphabet A - Z
and the letter b for “space”.

How many more letters can we add to our alphabet without any
further changes to our encoding procedure?

Would this just be 26 x 2 = 52
52 - 49 = 3, hence 3 more letters could be added?
I think this question is essentially asking, find the largest number of x's such that $x \rightarrow 9x \, mod \, 49$ is a bijection. Since $(9,49) = 1, 9\mathbb{Z}_{49} = \mathbb{Z}_{49}$.