5x ≡ 3 (mod 12).
The question asked me to find the smallest positive non zero integers that can be plugged in the place of x.
I only know how to find the answer by making a list.
5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 0, 5, 10, 3.
It took 12 for 3 to get back to another 3. So the answer is 3, 15(=3+12), 27(=3+12+12).
But is there any shortcut or formula to get them more easily?
Use Euclidean algorithm to find the multiplicative inverse,
$$12=2(5)+2$$
$$5=2(2)+1$$
Hence, we have $$1=5-2(2)=5-2(12-2(5))=5(5)-2(12)$$
Hence $$1 \equiv 5(5) \pmod{12}$$
$$5^{-1} \equiv 5 \pmod{12}$$
$$5x \equiv 3 \pmod{12}$$
Multiplying by the $5^{-1}=5$ on both sides,
$$x\equiv 5(3)\equiv 15 \equiv 3 \pmod{12}$$
$5\cdot5=25\equiv1\operatorname{mod}12$, so multiplying both sides of the congruence by $5$ gives $x\equiv15\equiv3\operatorname{mod}12$.
$5$ and $12$ are coprime, so by Bezout some linear combination is $1$. Indeed, $5\cdot5+(-2)12=1$. Thus, $5^{-1}\equiv5\bmod{12}$.
So solve the congruence by multiplying by $5^{-1}\equiv5$ on both sides. Get $x\equiv3\bmod{12}$.