Claim: A number is divisible by $4$ if and only if the number formed by the last two digits is divisible by $4$.
Here's where I've gotten so far.
Let $x$ be an $(n+1)$-digit number. So $x= a_na_{n-1} \dots a_2a_1a_0$. If $a_1 = 0$ and $a_0 =0$, then $x$ is a multiple of $100$ and therefore clearly divisible by $4$. So we must deal with the case when $(a_1 \neq 0 \lor a_0 \neq 0)$.
Then if $10a_1 + a_0 \equiv 0 \mod 4$ is true, then $x$ is divisible by $4$.
Do I need to do anything else or is this done? I feel like it's not quite complete, but I'm not sure how to proceed.
Answer
Pick $h, j \in \{0,1,\dots 9\}$ then $$100k + 10h + j \equiv 10h+j \mod 4$$ because $4 \mid 100$
So we have $$100k + 10h + j \equiv 0 \mod 4 \ \Leftrightarrow \ 10h + j \equiv 0 \mod 4$$
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