Divisible rules of 4

My proof for the challenge master's statement. IF YOU DON'T WANT TO SPOIL THE PROBLEM FOR YOURSELF, DON'T READ!!! $m,n,k \in \mathbb{N}$ $n := \overline{ d_{ m+k }d_{ m+k-1 }...d_{ m+1 }d_{ m }d_{ m-1 }...d_{ 0 } }$ $2^{m} \in \mathbb{N} | \overline{ d_{m-1}...d_{0} } \in \mathbb{N}$ $n- \overline{ d_{m-1} ... d_{0} } =\overline{ d_{m+k}d_{m+k-1}...d_{m+1}d_{m}0_{m \: times} } \in \mathbb{N}$ $2^{m} | n \Leftrightarrow 2^{m} | \overline{ d_{m+k}d_{m+k-1}...d_{m+1}d_{m}0_{m \: times} }$ $b := \overline{ d_{m+k}d_{m+k-1}...d_{m+1}d_{m} } \in \mathbb{N}$ $b \times 10^{m} = \overline{ d_{m+k}d_{m+k-1}...d_{m+1}d_{m}0_{m \: times} }$ $\frac{b \times 10^{m}}{2^{m}} = \frac{b \times 5^{m} \times 2^{m}}{2^{m}} = b \times 5^{m} \in \mathbb{N} \subset \mathbb{Z} \therefore 2^{m}|\overline{ d_{m+k}d_{m+k-1}...d_{m+1}d_{m}0_{m \: times} } \therefore \boxed{2^{m} | n}$ $Q.E.D.$

Easier proof of the challenge: take the number minus the last n digits. It has n trailing zeroes, and is divisible by 10^n, and thus is divisible by 2^n. Thus, the divisibility of the original number by 2^n depends on the divisibility of the last n digits by 2^n.

The solution above and the comments below explain exactly how to prove that this number is divisible by four. I merely observed that it ends in 12, which means that dividing by two gives a number that ends either in 56 or 06 (in this specific case actually 06), and is therefore itself divisible by two, meaning that the original number is divisible by four.

Why 15 is divisible by 5?

Then why is 100 divisible by 4?

So 14 divisible by 4?