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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?

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    and @arturo, that worked well than the answer given below although both intend to mean the same. thx guys2011-11-13

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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).