$12x \equiv 33 \pmod{57}$

Question

Can you show me a step by step instruction how to solve this problem?

$$12x \equiv 33 \pmod{57}$$

Answer 1

$$\begin{array}{rcll} 12x &\equiv& 33 & \pmod {57} \\ 4x &\equiv& 11 & \pmod {19} \\ 4x &\equiv& -8 & \pmod {19} \\ \end{array}$$

You can reduce them by $4$ as $4$ is relatively prime wiht $19$

$$\begin{array}{rcll} 4x &\equiv& -8 & \pmod {19} \\ x &\equiv& -2 & \pmod {19} \\ x &\equiv& 17 & \pmod {19} \\ \end{array}$$

$$x=19n+17$$

Answer 2

$$\begin{array}{rcll} 12x &\equiv& 33 & \pmod {57} \\ 4x &\equiv& 11 & \pmod {19} \\ (4x)(5) &\equiv& (11)(5) & \pmod {19} \\ 20x &\equiv& 17 & \pmod {19} \\ x &\equiv& 17 & \pmod {19} \\ x &\equiv& 17,36,55 & \pmod {57} \\ \end{array}$$

Answer 3

Hint:

Note first that it is equivalent to $$4x\equiv 11\pmod{19},$$ and to solve this equation, all you have to do is finding the inverse of $4\bmod 19$.

Answer 4

Hint $\ $ As $ $ explained here: $\quad \begin{align} & \bmod 19(\color{#c00}{3})\!:\,\ x\equiv \dfrac{11}{4}\color{#c00}{\dfrac{3}{3}}\\[.3em] \overset{\large\ \ \div\, \color{#c00}3}\iff\, &\bmod 19\!:\quad\ \ \ x\equiv \dfrac{11}{4}\equiv\dfrac{-8}{4}\equiv -2\equiv 17\end{align} $