# A divisibility rule for 19

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.

• I think you meant "if $\;10a+2b\;$ is divisible by $\;19\;$" – DonAntonio Aug 20 '16 at 20:14
• Sorry, you're only interested in two digit multiples of $19$? But there are only $5$ of them....$\{19,38,57,76,95\}$ and it is easy to check that in each case $a+2b=19$. Or did you mean something else? – lulu Aug 20 '16 at 20:33
• @DonAntonio No. See one of the answers. – user236182 Aug 20 '16 at 20:36
• @lulu Why two digits? Every natural number can be wrriten as $\;10a+b\;$ , with $\;a,b\in\Bbb Z\;$ . In fact, $\;b\;$ can be taken to simply be a digit and, in my opinion, it'd be way clearer had the OP written that. – DonAntonio Aug 20 '16 at 20:37
• @DonAntonio Oh, I took $a,b$ to be the digits of the number. So $10a+b$ equaled the original. On rereading, your interpretation is certainly better. – lulu Aug 20 '16 at 20:38

$10a+b$ is divisible by $19$ if and only if $20a+2b$ is divisible by $19$, of course $20a+2b\equiv a+2b\bmod 19$
• $20-1=19{}{}{}{}$ – Jorge Fernández Hidalgo Aug 21 '16 at 1:09