# How to solve the congruence equation $6x+y \equiv 19 \pmod {26}$?

I have the congruence equation:

$$6x+y \equiv 19 \pmod{26}$$

One way to solve it is to start from:

$$y \equiv 19-6x \space \pmod{26}$$

and try all the $$y\in \left \{ 0,\dots,25 \right \}$$. However my book states that there exist only $$12$$ possibilities for $$x$$, so I think the congruence equation $$6x+y \equiv 19 \pmod{26}$$ can be simplified. However I didn't succeed in doing it, can you help me?

$$6x+y-19$$ is to be divisible by $$26$$

hence $$6x+y\equiv19\pmod2$$

$$\implies y\equiv1\pmod2$$

If $$y=2z+1, 6x+2z+1\equiv19\pmod{26}$$

$$\iff z\equiv9-3x\pmod{13}$$

• So, there are $12+1$ possibilities – lab bhattacharjee May 2 '19 at 14:54
• I'm very sorry but I started today studying modular arithmetic and I don't understand the passages of your calculations, can you add more details? Thanks so much – AleWolf May 2 '19 at 15:15
• @Ale, we can write $$6x+y-19=26k\iff y=2(13k-3x+9)+1$$ for some integer $k$ – lab bhattacharjee May 2 '19 at 15:22
• Thanks so much, very clear now :) – AleWolf May 2 '19 at 15:36
• @Ale, when the modulus is composite, try taking modulus of the prime factors & their powers – lab bhattacharjee May 2 '19 at 20:06