...solution of a congruence with modulo notation from the reciprocal
Hey guys, new to the forum. I have been looking through my notes and I am stuck on an area of my number theory module.
I understand the modulo notation and for example that 14 ≡ 2 mod 12, etc etc.
I have the reciprocals (modulo 15):
1bar = 1
2bar = 8
3bar = No such x, also 5bar, 6bar, 9bar, 10bar, 12bar do not exist
4bar = 4
7bar = 13
8bar = 2
11bar = 11
13bar = 7
14bar = 14
I know this is just the reverse process but if someone could explain to me step by step how these solutions are found that would be great.