Having trouble with Chinese Remainder Theorem

Question

I am having trouble with the Chinese Remainder Theorem.

For this question..the equation $5x\equiv 3 \pmod6$ I found there is exactly one incongruent solution modulo $6$. But then I found 3 solutions modulo $6$?

Similarly, in this system

$$\begin{cases} x\equiv3 \pmod4\\ x\equiv5 \pmod3 \end{cases}$$

I am having trouble determining how many solutions there are? I think there is only one solution modulo $12$, but I am not so sure right now.

Answer 1

In both cases there is exactly 1 solution.

Easiest way to solve the second system is: $$ \begin{cases} x \equiv 3 \pmod 4 \\ x \equiv 5 \pmod 3 \end{cases} \Rightarrow \begin{cases} x = 3 + 4k \\ 3 + 4k \equiv 5 \pmod 3 \end{cases}\Rightarrow \begin{cases} x = 3 + 4k \\ k \equiv 2 \pmod 3 \end{cases}$$

$$ \Rightarrow\begin{cases} x = 3 + 4k \\ k = 2 +3l \end{cases} \Rightarrow x = 3 + 4(2+3l) = 11 + 12l$$

So: $$x \equiv -1 \pmod{12}$$

Answer 2

\begin{align} x \equiv 3 \pmod 4\\ x \equiv 5 \pmod 3 \end{align}

For really small number such as this, you don't need a lot of high-powered math to get a feel for what the solution should look like

$x \equiv 3 \pmod 4 \implies x \in \{\dots, 3, 7, \color{red}{11}, 15, 19, \color{red}{23}, 27, 31, \color{red}{35}, 39, \dots\}$

$x \equiv 5 \pmod 3 \implies x \in \{\dots, 2, 8, \color{red}{11}, 14, 17, 20, \color{red}{23}, 26, 29, 32, \color{red}{35}, 38 \dots\}$

You can see that $x \in \{\dots, 11, 23, 35, \dots\}$

Which doesn't prove but strongly suggests that $x \equiv 11 \pmod{12}$