How we can prove divisible rules for larger primes like $73$?
$n$ is dividable by $73$ if and only if $73|a-b$ where $a$ is number from first $4$ digits $b$ is number from rest of digits
example $$ 43181169 \rightarrow 4318 -1169 = 3149 = 47 \cdot 67 $$
$\begin{align}{\bf Hint}\ \ \bmod 73\!:\,\ \color{#c00}{10^{\large 4}\equiv -1}\ \Rightarrow\ &\ d_0\,+\, d_1 \color{#c00}{10^{\large 4}} + d_2 \color{#c00}{10^{\large 8}}+\ \cdots\ \ \text{(in radix } 10^{\large 4}\ \text{with digits } d_i )\\[.2em] \equiv\ &\ d_0\, -\, d_1\ \ \ +\ \ \ d_2\ \ \ \ \ \ -\ \cdots\ \ \text{(alternating digit sum)}\end{align}$
Same mod $\,137\,$ by $\, 10^{\large 4}+1 = 73\cdot 137.\ $
If we consider an integer in radix $\, 10^{\large 4}$ as a polynomial $\,P(10^{\large 4})$ in the radix then above is
$\!\bmod\, 10^{\large 4}+1\!:\,\ \color{#c00}{10^{\large 4}\equiv -1}\,\Rightarrow\, P(\color{#c00}{10^{\large 4}})\equiv P(\color{#c00}{-1}) \equiv\, $ alternating sum of digits in radix $10^{\large 4} $
which is the radix $10^{\large 4}$ analog of casting $11$'s in radix $10,\,$ i.e. the common test for divsibility by $11$, where the above inference employs the Polynomial Congruence Rule.
Remark $ $ The same method works if we replace $\,73\,$ by any integer $\,n\,$ coprime to $\,10\,$ since then $\,10^{\large k}\equiv 1 \pmod{\!n}$ for some integer $\,k>1,\,$ e.g. we can choose $\, k = \phi(n)\,$ by Euler's Theorem. The least such $\,k\,$ is known as the order of $10$ modulo $\,n,\,$ and it must divide $\,\phi(n).$
See also the closely related topic of periodicity of decimal expansion of rationals (fractions). e.g. $1/73\, =\, 0.\overline{01369863}\,$ repeats with period $\,\color{#0a0}8,\,$ and $\,0136 + 9863 = 9999,\,$ because $\!\bmod 73\!:\,\ \color{#c00}{10^{\large 4}\equiv -1}\,\Rightarrow\, 10^{\large\color{#0a0} 8}\equiv 1\,$ so $\,10\,$ has order $\,\color{#0a0}8\,$ modulo $73,\,$ by the Order Test.
If $a\equiv b\pmod{p}$, then you have $a^k\equiv b^k\pmod{p}$. So, for any polynomial with coeffs. in $\mathbb{Z}$, you can say that $$f(a)\equiv f(b)\pmod{p}$$ here, consider the polynomial $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n $. So, $f(10)$ is the number $a_n\cdots a_1a_0$.
As, $10^4\equiv -1\pmod{73}$, we have $f(10^4)=f(-1)$. That means, \begin{align} f(10)& \equiv \overline{a_na_{n-1}\cdots a_1a_0}\\ & \equiv \overline{a_3a_2a_1a_0}+\overline{a_7a_6a_5a_4}\cdot 10^4 + \overline{a_{11}a_{10}a_9a_8}\cdot 10^8+\cdots \\ & \equiv \overline{a_3a_2a_1a_0}-\overline{a_7a_6a_5a_4}+\cdots +(-1)^{\lfloor n/4 \rfloor}\overline{a_na_{n-1}a_{n-2}a_{n-3}}\pmod{73} \end{align} is the remainder when you devide the number by $73$.
In general, if possible, find $k$ such that $10^k\equiv \pm 1\pmod{p}$ and you will have $f(10^k)\equiv f(1)$ or $f(-1)\pmod{p}$, using this property you can check the divisibility. Here you can get some example.