Prove $4|n$ if and only if $4|(10a_1+a_0)$

Question

Prove $4|n$ if and only if $4|(10a_1+a_0)$. I know there may be "duplicates" out there, but most of them are for the case of $3|n$ and stuff like that. I'm still confused as to how the thought process works though. I have to prove the statement for $n=(a_k \times 10^k) + (a_{k-1} \times 10^{k-1}) + \cdots + (a_1 \times 10^1) + (a_0 \times 10^0)$. Before having to prove this in my homework assignment, I had to prove that if $n$ is an integer and $n \geq 2$, then $10^n \equiv 0 (mod 4)$.

Answer 1

Your idea is good. If we can show that $10^n\equiv0$ modulo $4$ for "most" $n$ ($n>1$), then the problem is done already. Now notice that $4\times 25=100=10^2$, and $10^{n}=10^210^{n-2}$ for $n\geq 2$. Can you continue?

Answer 2

\begin{align} n&=(a_k \times 10^k) + (a_{k-1} \times 10^{k-1}) + \cdots + a_1\times 10^2+(a_1 \times 10) + a_0\\ &= \underbrace{(a_k \times 10^{k-2}) + (a_{k-1} \times 10^{k-3}) + \cdots + a_1)\times 10^2}_{m} + (a_1 \times 10) + a_0 \end{align} so $\;n=m+10a_1+a_0$, and $m$ is divisible by $4$. Can you proceed?