<s>[INST] Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other,  but got nothing. No idea what to do.
 [/INST] $$\overline{xyz138}=1000\overline{xyz}+138\equiv 6\overline{xyz}-2\pmod 7\\ \overline{138xyz}=138000+\overline{xyz}\equiv 5+\overline{xyz}\pmod{13}\\ \overline{x1y3z8}\equiv 10x+1+10y+3+10z+8\equiv 1-x-y-z\pmod{11}$$
Denote $a:=\overline{xyz}$, then you have the linear system
$$\begin{cases}6a-2\equiv 0\pmod 7\\5+a\equiv 6\pmod{13}\end{cases}$$
along with $x+y+z\equiv 7\pmod{11}$
Apply CRT to solve modulo $7\times 13$ and then check the values of $a$ that satisfy the congruence modulo $7\times 13$ against the final constraint. There are only $11$ values of $a$ to check.
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