law of divisibility on $37$ how to find and prove law of divisibility on $37$?
Thanks in advance.
Added:---- how to prove for$37$ that: Split off the last digit, multiply by 11, and subtract the product from the number that is left. if the result is divisible by 37 then so is the original number.divisibility criteria for 37
 A: As $$999=37\cdot27$$
So, $10^3\equiv1\pmod{37}\implies 10^{3k}\equiv1\pmod{37}$
So, $\sum_{0\le r\le n}a_i10^i$
$$=(a_0+10a_1+100a_2)+10^3(a_3+10a_4+100a_5)+10^6(a_6+10a_7+100a_8)\cdots$$
$$\equiv (a_0+10a_1+100a_2)+(a_3+10a_4+100a_5)+(a_6+10a_7+100a_8)\cdots\pmod {37}$$
i.e., we can group by $3$ digits and add to test the divisibility by $37$ 
EDIT: the explanation of the link in the Question
As $11\cdot(10a+b)-1\cdot(11b-a)=111\cdot a$
$(10a+b)$ will be divisible by $111\iff (11b-a)$ is  divisible by $111$
A more general idea can be found here.
Try to find how  $11,1$ are identified as multiplier from the link.
A: HINT: $37$ divides $999=37\cdot27$, so $37\equiv-1\pmod{1000}$. Now think about the classic divisibility test for $9\equiv-1\pmod{10}$. Imagine writing the number to be tested in base $1000$.
A: $\rm mod\ 37\!:\ 100^{-1}\!\equiv 10\:$ so $ $ e.g. $\rm\:37\mid  n = 100^2 c + 100 b + a \iff 37\mid 10^2 n \equiv c + 10 b + 10^2 a$
$37\mid \color{#0a0}{54}\color{blue}{39}\!:\quad \color{#0a0}{54}\!+\!10(\color{blue}{39}) \equiv \color{0a0}{17}\!+\!10(\color{blue}2)\equiv 0 $
$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{44}\color{blue}{47}\color{#c00}{40}\!: &&\!\!\color{#0a0}{44}\!+\!10(\color{blue}{47}\!+\!10(\color{#c00}{40}))\\ \equiv && \color{#0a0}7\!+\!10(\color{blue}{10}\!+\!10(\color{#c00}3))\\ \equiv && \color{#0a0}7\!+\!10(3)\equiv 0\end{eqnarray}$
$\rm\!\begin{eqnarray} 37\mid \color{#0a0}{54}\color{blue}{66}\color{#c00}{65}38\!:  &&\color{#0a0}{54}\!+\!10(\color{blue}{66}\!+\!10(\color{#c00}{65}\!+\!10(38))) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10(\color{#c00}{28}\!+\!10)) \\ \equiv && \color{#0a0}{17}\!+\!10(\color{blue}{29}\!+\!10)\\
\equiv &&\color{#0a0}{17}\!+\!10(2)\equiv 0 \end{eqnarray}$
A: Given a number in base 10, say $d_nd_{n-1}\cdots d_0$, calculate $d_0 + 10d_1+26d_2+d_3+10d4+26d_5+\cdots$ (alternating $1$, $10$, and $26$). The resulting number is divisible by 37 if and only if $d_nd_{n-1}\cdots d_0$ is divisible by $37$.
