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.

  • $\begingroup$ I think you meant "if $\;10a+2b\;$ is divisible by $\;19\;$" $\endgroup$ – DonAntonio Aug 20 '16 at 20:14
  • $\begingroup$ 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? $\endgroup$ – lulu Aug 20 '16 at 20:33
  • $\begingroup$ @DonAntonio No. See one of the answers. $\endgroup$ – user236182 Aug 20 '16 at 20:36
  • $\begingroup$ @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. $\endgroup$ – DonAntonio Aug 20 '16 at 20:37
  • $\begingroup$ @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. $\endgroup$ – 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 $20a+2b\equiv a+2b\bmod 19$

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  • $\begingroup$ How can the last step be justified? $\endgroup$ – IronMan12 Aug 20 '16 at 21:37
  • $\begingroup$ $20-1=19{}{}{}{}$ $\endgroup$ – Jorge Fernández-Hidalgo Aug 21 '16 at 1:09

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