What is the best method to solve congruence equations??

Question

I am trying to learn congruence equations for discrete mathematics but this concept is actually giving me a lot of issues. In this instance:

$$15x + 12 \equiv −11 (\text{mod }49)$$

And so I tried it like this:

$$15x \equiv −23 (\text{mod }49)\\ 15x \equiv 26 (\text{mod }49)\\ 1 = 49(4)+15(-13)$$ (From Euclidean Algorithm)

From here I don't actually know what to do anymore. I know I need the inverse modulo n but I don't know which one it is, and I also don't know what to do with it anyway. Can anyone help me figure out how to solve these equations?

See the linked dupes for most all common methods for solving linear congruences, e.g. $!\bmod 49!:\ 15x \equiv -23 \iff x \equiv \dfrac{-23}{15} \underset{\times\ 3}\equiv \dfrac{-20}{-4} \equiv 5,$ by Gauss's algorithm, using least magnitude remainders. — Bill Dubuque, Apr 23 '23 at 23:54

score 0 · Answer 1 · answered Apr 23 '23 at 23:03

0

The equation $1 = 49(4)+15(-13)$ tells you that $15*(-13)$ is the same as $1$ (modulo $49$).

Therefore the equation you obtained by multiplying by $-13$ tells you that $x ≡ 26 * (-13)$(mod $49$).

answered Apr 23 '23 at 23:03

user1172706

1,405

Does that mean that x≡26∗(−13) (mod 49) is the final answer? – James Apr 23 '23 at 23:05
Simply multiply out 26 and -13 mod 49. – user1172706 Apr 23 '23 at 23:06
x≡-338 (mod 49) which reduces to x≡5 (mod 49)! – James Apr 23 '23 at 23:14
Yes, that's right – user1172706 Apr 23 '23 at 23:18
oh ok the concept of 15∗(−13) basically equaling 1 was messing me up – James Apr 23 '23 at 23:19

What is the best method to solve congruence equations??

1 Answers1