General Proof of Divisibility Tests

 Statement:  A number is divisible by $4$ if the last $2$-digits are divisible by $4$. 

Proof.

Let $n=a_m 100^m + a_{m-1} 100^{m-1}+ \cdots+a_2 100^2 + a_1 100 + a_0$.   be an integer, 

where $a_k$  are integers and $0\leq a_k\leq 99, k=0,1,\dots,m$
Let us consider  the polynomial

$f(x)=a_m x^m + a_{m-1} x^{m-1}+ \cdots+a_2 x^2 + a_1 x + a_0$.

We have

    $100 \equiv 0(\mod 4)$

$\Rightarrow f(100) \equiv f(0)(\mod 4)$

 $\Rightarrow n \equiv a_0 (\mod 4)$  ( $\because f(100)=n$  and  $f(0)=a_0$)

 $\Rightarrow 4|(n-a_0)$.

 $\Rightarrow n-a_0=4q$  for some $q\in\mathbb{Z}$

$\Rightarrow n=a_0+4q$

Hence  $4|n$ iff $4|a_0$.

For example :  $34,164$ is divisible by $ 4$  because $64$  is divisible by $4$.