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.
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$\begingroup$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!
$\endgroup$ $\begingroup$You want to prove that an integer is divisible by $4$ if and only if the number formed from its last two digits is divisible by $4$. Observe that $a=10^n x_n + \cdots + 10 x_1 + x_0$ can be written as $100 b + (10 x_1 + x_0)$. Because $100b$ is a multiple of $4$, we have that $a$ is a multiple of $4$ if and only if $10x_1+x_0$ is a multiple of $4$.
In your proof, the expression $2(2k+5)(b)$ can be written as $4k+2 \cdot 5 \cdot b$, where $b$ is of a sum of terms each of which is divisible by $10$. So you can pull out this $10$ and you get $2 \cdot 5 \cdot 10 = 100$. Thus, your expression is a sum of $4k$ and a multiple of $100$, as desired.
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