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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.

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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 $\{1+ 3k \ | \ k\in\mathbb Z \}$.

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    $\begingroup$ I suggest suppressing the $\times$ symbols or replace them with \cdot; it will make your answer more readable. $\endgroup$ – Simon S Nov 14 '15 at 17:21
  • $\begingroup$ @Simon S : I did it, I did not know what convention would be better. Thank you ! $\endgroup$ – Balloon Nov 14 '15 at 17:24
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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 $\overline{1}$.

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

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