Tough Divisibility Problem 
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$. 

$$2A13B \equiv 12 \pmod{19}$$
$$20000 + 1000A + 100 + 30 + B \equiv 12 \pmod{19}$$ 
$$ 5 + 12A + 5 + 11 + B \equiv 12 \pmod{19}$$
$$  21+ 12A+ B \equiv 12 \pmod{19}$$
$$ 12A+ B + 9 \equiv 0 \pmod{19}$$
This is where I'm stuck.

\begin{align}30\,000+1\,000A+200+10+B\equiv x\pmod{19}&\iff-1+12A+10+10+B\equiv x\pmod{19}\\&\iff12A+B\equiv x\pmod{19}.\end{align}Therefore, since $12A+B+9\equiv0\pmod{19}$, take $x=10$.
