Divisibility by 4 I was asked to find divisibility tests for 2,3, and 4.
I could do this  for 2 and 3,  but for 4.
I could come only as far as:
let $a_na_{n-1}\cdots a_1a_0$ be the $n$ digit number.
Now from the hundredth digit onwards, the number is divisible by 4 when we express it as sum of digits.
So, the  only part of the proof that's left  is to prove that $10a_1+a_0$ is divisible by 4.
So if we show that this happens only when the number $a_1a_0$ is divisible by 4, the proof is complete.
So the best way to show it is by just taking all combinations of $a_1,a_0$ or is there a better way?

Note that $10a_1+a_0\equiv2a_1+a_0$ (mod 4). So for divisibility by 4, $a_0$ must be even and in this case $2a_1+a_0=2(a_1+\frac{a_0}{2})$. So, $a_1$ and $\frac{a0}{2}$ must be of same parity (means both are either even, or odd).
