Find all integer solutions to $4x \equiv 13 \mod 3$

Question

I am confused on how to go about finding all integer solutions of $$ 4x \equiv 13 \mod 3$$

I am not understanding how to solve congruence problems and I am thus looking for an explanation how to solves this problem.

Answer 1

You have that $$4\cdot x\equiv 13\,[3]\iff(3+1)\cdot x\equiv3\cdot4+1\,[3]\iff x\equiv 1\,[3]$$ and then the solutions are $\{1+ 3k \ | \ k\in\mathbb Z \}$.

Answer 2

we have $x\equiv \frac{13}{4}\equiv \frac{16}{4}\equiv 4\mod 3$

Answer 3

This congruence is actually pretty simple, as $4 \equiv 1 \pmod{3}$, and $13 \equiv 1 \pmod{3}$ as well.

Hence $$x\equiv1\pmod{3}$$ which gives solution $\{1+3k\mid k\in\mathbb{Z}\}$.

Edit: You can also look at them as equivalence classes of the quotient $\mathbb{Z}/3\mathbb{Z}$. Since class multiplication $\overline{x}\cdot\overline{y}=\overline{xy}$ is well-defined: $$\overline{4x} = \overline{13}$$ $$\overline{4}\overline{x} = \overline{13}$$ $$\overline{1}\overline{x} = \overline{1}$$ Therefore, $x$ must be a representative of $\overline{1}$.