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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.

IronMan12
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  • 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
  • @lulu $b$ is the last digit of a positive integer and $a$ is the "remaining leading truncated number". E.g., in $194$ we would get $b=4$, $a=19$. – user236182 Aug 20 '16 at 20:39
  • @user236182 I can't see why the unique answer there is answers that. Besides, the OP's words are "add two times the last digit to the remainded leading truncated number..." If the number is $;10a+b;,;;b;$ a digit, then $;2b;$ must be added to $;10a;$, not to $;a;$ ...or I am misunderstanding something. – DonAntonio Aug 20 '16 at 20:39
  • @DonAntonio For contradiction, let your statement be true. If $19\mid 10a+2b$, then $19\mid 10a+b$. But then $19\mid (10a+2b)-(10a+b)=b$, so $b$ is divisible by $19$. If $b$ is taken as a one-digit number, then $b=0$. So it wouldn't be true for a number with a different last digit than $0$, contradiction. – user236182 Aug 20 '16 at 20:41
  • @user236182 True, I fell into my own explanation contradiction...if $;10a+b;$ is a number, with $;a\in\Bbb Z,,,,b;$ a digit, then it surely must be $;a+2b;$. Thanks. – DonAntonio Aug 20 '16 at 20:42
  • By the general divisibility test in the linked dupe: $$ \color{#0a0}{19}\mid \color{#c00}{10},a+b !\iff! 19\mid a+\color{#c00}c,b,\ \ {\rm for}\ \ , \color{#c00}{c\equiv \dfrac{1}{10}}\equiv\dfrac{2}{20}\equiv\color{#c00}{\dfrac{2}{1}}!!!\pmod{!\color{#0a0}{19}}\qquad$$ – Bill Dubuque Nov 27 '23 at 18:35

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

Asinomás
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