Divisibility by Power of 2

Theorem
Let $r \in \Z_{\ge 1}$ be a strictly positive integer.

An integer $N$ expressed in decimal notation is divisible by $2^r$ the last $r$ digitd of $N$ forms an integer divisible by $2^r$.

That is:
 * $N = [a_0 a_1 a_2 \ldots a_n]_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $2^r$


 * $a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r$ is divisible by $2$.
 * $a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r$ is divisible by $2$.

Proof
First note that:
 * $10^r = 2^r 5^r$

and so:
 * $2^r \mathrel \backslash 10^r$

where $\backslash$ denotes divisibility.

Thus:
 * $\forall s \in \Z: s \ge r: 2^r \mathrel \backslash 10^s$

but:
 * $\forall s \in \Z: s < r: 2^r \nmid 10^s$

Thus let $N$ be divisible by $2^r$.

Then:

Hence the result.