<s>[INST] Having problem to find all solutions using the Chinese Remainder Theorem Find all solutions using the Chinese Remainder Theorem.
x = 2 (mod 4)
x = 3 (mod 5)
x = 9 (mod 13)

My step is here:
a = 2 (mod 4) , a = 0 (mod 5) , a = 0 (mod 13)
b = 0 (mod 4) , b = 3 (mod 5) , b = 0 (mod 13)
c = 0 (mod 4) , c = 0 (mod 5) , c = 9 (mod 13)

The general solution for x is given by x = a + b + c + k*lcm(4,5,13) = a+b+c+260k
since a = 0 (mod 5) and a = 0 (mod 13), so a = 65m for some integer m
then 65m = 2 (mod 4)
What should i do next?
I think the step i write 65m - 2 = 4 is wrong
 [/INST] We have , 
$$x\equiv2 \mod 4 \implies \color{#d05}{x=4a+2}\\ $$ 
$$\begin{align}x&\equiv3 \mod 5 \\4a+2&\equiv3\mod5\\  4a&\equiv 1\mod5 \\ a &\equiv 4 \mod 5\implies\color{#2cd}{a = 5b+4}\end{align} $$ 
$$\begin{align}x&\equiv9 \mod 13 \\ 4(5b+4)+2&\equiv 9\mod13 \\ 20b+18&\equiv9\mod13 \\ 20b&\equiv 4\mod 13 \implies  \color{#2c0}{b = 13c+8}\end{align}  $$ 
So , $x = 4a+2 = 20b+18=260c + 178$
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