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.
 A: Hint : $B \equiv -12A-9 \pmod{19}$. Next, in $3A21B \pmod{19}$ you can replace $B$ by the RHS expression for it.
A: \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 $x=10$.
A: You have some errors, which I will fix:
$$\begin{align}\overline{2A13B}=20000 + 1000A + 100 + 30 + B &\equiv 12 \pmod{19} \Rightarrow \\
(19\cdot 1052+12)+(19\cdot 52+12)A+(19\cdot 5+5)+(19\cdot 1+11)+B&\equiv 12 \pmod{19} \Rightarrow \\
12+12A+5+11+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+19\cdot 1+9+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+9+B&\equiv 12 \pmod{19} \Rightarrow \\
12A+B&\equiv 3\pmod{19}.\end{align}$$
Since $0\le A,B\le 9$, then: $(A,B)=(0,3),(3,5),(6,7),(9,9)$.
Similarly:
$$\begin{align}\overline{3A21B}=30000 + 1000A + 200 + 10 + B &\equiv x \pmod{19} \Rightarrow \\
(19\cdot 1578+18)+(19\cdot 52+12)A+(19\cdot 10+10)+10+B&\equiv x \pmod{19} \Rightarrow \\
18+12A+10+10+B&\equiv x \pmod{19} \Rightarrow \\
12A+19\cdot 2+B&\equiv x \pmod{19} \Rightarrow \\
12A+B&\equiv x \pmod{19}.\end{align}$$
So, $x=3$.
For example, take $\overline{2A13B}=20133\equiv 12 \pmod{19}$ and $\overline{3A21B}=30213\equiv 3\pmod{19}$.
A: Hint: $3A21B = 2A13B + 10000 + 80$
