A divisibility rule for 19

Question

Proof the following divisibility test for 19:

Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number.

More mathematically:

Let $a, b \in \mathbb{Z}$. Proof that $10a+b$ is divisible by 19 if $a+2b$ is divisible by 19.

My guess is that we can proof this using congruences.

Answer 1

$10a+b$ is divisible by $19$ if and only if $20a+2b$ is divisible by $19$, of course $20a+2b\equiv a+2b\bmod 19$