Verification of proof for when a number is divisible by 4 I have never taken a number theory course and so am only going off of the first few chapters in an introductory number theory book. The divisibility property I wish to prove is the following: 
Define an integer $$a=10^nx_n+10^{n-1}x_{n-1}+\cdots+10x_1+x_0$$ with all $x_i\in\{0,1,2,3,4,5,6,7,8,9\}$. If $$10x_1+x_0\equiv0\ (\text{mod}\ 4),$$ then $$a\equiv0\ (\text{mod}\ 4). $$ 
My attempt:
By definition $10x_1+x_0\equiv0\ (\text{mod}\ 4)\implies 10x_1+x_0=4k,$ with $k\in \mathbb{N}.$ Adding the necessary terms yields \begin{align}10^nx_n+10^{n-1}x_{n-1}+\cdots+10x_1+x_0&= 4k+10^nx_n+10^{n-1}x_{n-1}+\cdots+10^2x_2\\ &=2\big(2k+5(10^{n-1}x_n+10^{n-2}x_{n-1}+\cdots+10x_2)\big). \end{align} Therefore, $$a\equiv0\ (\text{mod}\ 2). $$ We define a number $Q\in\mathbb{N}$ such that $$2\big(2k+5(10^{n-1}x_n+10^{n-2}x_{n-1}+\cdots+10x_2)\big)\equiv Q\ (\text{mod}\ 4) $$ And this is where I get stuck. The result I'm after will follow if I could prove that $Q=0$, but I don't know how to do that. Any tips/hints would be appreciated.

Let $a := a_0 + 10a_1  + \cdots + 10^n a_n $. Since $$10^2 = 4\cdot 25 \equiv 0 \mod 4,$$ you get 
$$
a = a_0 + 10a_1 + 10^2\cdot (a_2 + \cdots 10^{n-2}a_n) \equiv a_0 + 10a_1 \mod 4.
$$
So If $a_0 + 10a_1$ is divisible by $4$, then so is $a$.
Addendum: It is a pleasant exercise doing these divisibility criteria for other numbers!
