FAQ: Congruences: Lesson - Solving Linear Congruence Problems

This community-built FAQ covers the “Solving Linear Congruence Problems” exercise from the lesson “Congruences: Lesson”.

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I cant reach the solution with the provided material at all. Makes no sense to me

So 2 and 7 are congruent mod 5 because
2 % 5 = 2 and 7 % 5 = 2

details

so 2, 7, 12, 17, … would be in the same congruence class mod 5

and if I wanted to solve
3x + 2 ≡ 1 (mod 5)
like I’d solve a linear equation

I’d subtract 2 from both sides of the ≡
3x ≡ -1 (mod 5)
but I don’t like negative numbers,
so I can add 5 (or a multiple of 5) so that the stuff would be the same mod 5
3x ≡ 4 (mod 5)
I want to divide both sides of the ≡ by 3,
but 4 can’t be divided by 3 evenly [so that I have an integer]
so I can keep adding 5 until I get something I can divide by 3 on the right of the ≡
3x ≡ 9 (mod 5)
I can divide 9 by 3 evenly
so I divide both sides of the ≡ by 3
x ≡ 3 (mod 5)
so the solution for x is 3 or anything else that’s 3 (mod 5)