An interesting question about divisibility by $19$. I came across an intriguing problem about divisibility as follows:

Prove that no matter how many $3$'s are inserted between the two zeroes of $12008$, the new number generated will always be divisible by $19$.

I went like this:
Suppose there are $k$ $3$'s inserted. The new number equals
$120\cdot10^{k+2}+3\cdot(10^{k+1}+10^k+\cdots+10^2)+8\\
=120\cdot10^{k+2}+3\cdot\frac{10^2\cdot(1-10^k)}{1-10}+8\\
=120\cdot10^{k+2}+\frac{10^{k+2}-100}{3}+8\\
=\frac{361\cdot10^{k+2}-76}{3}\\
=19\cdot\frac{19\cdot10^{k+2}-4}{3}$
Clearly it should be a multiple of $19$ since the fraction part can easily be proven to be an integer using mod.
I think I did it right, but I also want to know if there exists any other kind of approach, such as induction or some intuitive thoughts, etc. Cleaner solutions and correction on my solution are also welcomed. Thank you!
This is such an interesting fact to me.
 A: There is an easy induction as  $$12008=19 \cdot 632$$ and  $$(120\cdot10^{k+3}+3\cdot(10^{k+2}+10^{k+1}+10^k+\cdots+10^2)+8) \\ -(120\cdot10^{k+2}+3\cdot(10^{k+1}+10^k+\cdots+10^2)+8) \\ = 120\cdot10^{k+3} + 3\cdot10^{k+2}- 120\cdot10^{k+2} \\= 1083 \cdot10^{k+2} \\= 19 \cdot 57 \cdot10^{k+2}$$
so you start with a multiple of $19$ and keep adding multiples of $19$
A: Yes, you can prove this by induction too.
If $A_n$ is the number which has $n$ threes between the two zeros, then it is easy to see that
$$A_{n+1} = \left(\frac{A_n - 8}{100} \cdot 10 + 3\right) \cdot 100 + 8$$
How do you see this? You see it constructively. Think what the first operation $\color{red}{-8}$ does to $A_n$. Then think what the second operation $\color{red}{/100}$ does and so on. You will see that I am really constructing $A_{n+1}$ starting from $A_n$.
If you simplify this you get:
$$A_{n+1} = 10 \cdot A_n + 19 \cdot 12$$
From the last equality it is obvious how one can do the induction.
A: Not sure it is easier, but you can do it inductively.
Let $$a_0=12008=632\times 19$$$$ a_n=100\left(\frac {a_{n-1}-8}{10}+3\right)+8=10(a_{n-1}-8)+308=10a_{n-1}+228$$
Noting that $228=19\times 12$ the induction is straight forward.
A: Let us find how $3$ has been identified
$$120\cdot10^{k+2}+a\sum_{r=2}^{k+1}10^r+8$$
$$\equiv6\cdot10^{k+2}+a\cdot\dfrac{10^2(10^k-1)}9+8$$
$$\equiv\dfrac{10^{k+2}(54+a)+72-100a}9$$
So, it is sufficient to have $$54+a\equiv0\pmod{19}\text{  and }100a\equiv72\pmod{19}$$
$$a\equiv-54\equiv3\text{ and } 5a\equiv-4\equiv15$$
$$\implies a\equiv 3\pmod{19}$$
For numbers  in base $10, 0\le a\le9$
A: $$\begin{align} n &\,=\,\ \  [a]33\cdots 3308\\[.1em]
\Longrightarrow\ 3n &\,=\, [3a]99\cdots9924\\[.1em]
&\,=\, 3a(10)^k\! \color{#c00}{+ 10^k}\!-76\\[.1em]
&\,\equiv\, (3a\!\color{#c00}{+\!1})10^k\!\!\!\pmod{\!19}\\[.2em]
{\rm so}\ \ 19\mid n&\iff 19\mid 3a\!+\!1\iff \underbrace{a\equiv 6\!\!\!\pmod{\!19}}_{\textstyle {\rm e.g.}\ \ \ a = 120}
\end{align}\qquad$$
