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I'm a beginner in number theory, I was trying to prove the divisibility rule of 19, can someone help me to complete my proof

Let $n$ be $10a + b$ where b is the unit digit and 10a is the rest; 10 is coprime to 19,
$n \equiv 0 \pmod{19} \iff a+2b \equiv 0 \pmod {19} \iff x\cdot a + x\cdot 2b \equiv 0 \pmod{19} \iff 10a + b \equiv 0 \pmod{19}$
Where it holds that $10a + b - (x\cdot a + x\cdot 2b) = 19k \space or \\ 10a + b - (x\cdot a + x\cdot 2b) \equiv 0 \pmod{19}$

but I have a problem finding the $x$ multiplier.

I've visited A divisibility rule for 19, Didn't quite understand what has been done there; and the important factor is that I want the approach I'm taking completed.

If my proof is completely meaningless etc, I'd appreciate a proof like it.
Thanks

BGOPC
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    Are you sure you want $xa + xb$? Typically they go for $a + xb$, so you're adding x times the last digit to the rest of the number. – Calvin Lin Nov 27 '23 at 16:16
  • Maybe, I tried to take the approach used for 17, and 13, I'm like really beginner – BGOPC Nov 27 '23 at 16:18
  • Sure, can you write up the approach used for 17? That will help us tell if you make an error trying to generalize the approach. – Calvin Lin Nov 27 '23 at 16:19
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    Welcome to Mathematics Stack Exchange. Try $x=10$. Cf. this question. And it's confusing to let $a$ be $10a+b$ – J. W. Tanner Nov 27 '23 at 16:19
  • @CalvinLin Proof the divisibility rule of 17 – BGOPC Nov 27 '23 at 16:20
  • (Apart from your $a$ doing double duty like Tanner pointed out) You see what I mean by "Typically they go for $a + xb$?" In the link, they say that $17 \mid a - 5 b \Leftrightarrow 17 \mid 10a + b$. – Calvin Lin Nov 27 '23 at 16:22
  • I edited the question so it isn't 2 a's in use You see the approach taken in the 17(as I understood) is that they're subtracting $x$ from $17k$ to get to $17j$ resulting in x also a multiply of 17 I tried to take the same approach But I don't know what Should be $x$ resulting in a $19j$ – BGOPC Nov 27 '23 at 16:30
  • As $2$ is the multiplicative inverse of $10$ modulo $19$, the result follows from this more general result. Not voting to close this as a duplicate yet, because A) I happened to answer that query and B) it is possible that's not the first incarnation on our site. – Jyrki Lahtonen Nov 27 '23 at 17:42
  • By the general divisibility test in the linked dupe: $$ \color{#0a0}{19}\mid \color{#c00}{10},a+b !\iff! 19\mid a+\color{#c00}c,b,\ \ {\rm for}\ \ , \color{#c00}{c\equiv \dfrac{1}{10}}\equiv\dfrac{2}{20}\equiv\color{#c00}{\dfrac{2}{1}}!!!\pmod{!\color{#0a0}{19}}\qquad$$ See the detailed proofs there (which are no simpler in this special case). $\ \ $ – Bill Dubuque Nov 27 '23 at 18:38

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Thanks to user J. W. Tanner. The answer for $x$ is 10, Giving us this $a+2b \equiv 0 \pmod {19} \iff 10a + 20b \equiv 0 \pmod{19} \iff 10a + b \equiv 0 \pmod{19} \iff n \equiv 0 \pmod{19}$
This is Because:
$10a + b - (10a + 20b) = -19b \equiv 0 \pmod{19} \iff x \equiv 0 \pmod{19}$

BGOPC
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    I think that after the second double implication, the expression should be $10a+20b$ (rather than $20a+20b$). – paw88789 Nov 27 '23 at 16:54
  • Oh, yes thanks i prob made a mistake while typing it – BGOPC Nov 27 '23 at 16:56
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    I think you don't want to begin your chain of implications with $n \equiv 0 \pmod{19}$ since that is what you are trying to prove. It's where the chain ends. – Ethan Bolker Nov 27 '23 at 17:00
  • I think you meant $10a+b-(10a+20b)=\color{red}-19b$, but it's still $\equiv0\pmod{19}$; and the choice $x=10$ was rather obvious when you wrote $a+2b\dots\iff 10a\dots$ in the question – J. W. Tanner Nov 27 '23 at 17:15
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Nov 27 '23 at 18:55