For a positive integer $n$ there exists an integer $a_{n-1}a_{n-2}...a_0$, where the $a_{i}'s $ take the values 1 and 2 and it is divisible by $2^n$ For example, for $n=2$, $m$ is $12$.
For $n=3$, $m=112$.
How to prove it for any integer $n$?
 A: Your examples show that
it is true for
$n=2, 3$.
Suppose it is true for $n$,
and
$2^n | A_n
= a_{n-1}a_{n-2}…a_0
$.
Consider
$A_{n+1}
= a_na_{n-1}a_{n-2}…a_0
= 10^n a_n +A_n
$.
Let
$B_n = \dfrac{A_n}{2^n}
$.
Then
$A_{n+1}
=10^na_n+2^nB_n
=2^n(5^na_n+B_n)
$.
To make
$2^{n+1} | A_{n+1}$,
it will be enough
if 
$5^na_n+B_n$
is even.
To do this, 
choose $a_n$
to have the same parity
as $B_n$
(i.e., even if $B_n$ is,
odd if not).
In particular,
we can choose
$a_n = 2+(B_n \bmod 2)$
so that,
starting with the examples
for $n=2, 3$,
we can generate
as many $a_n$ as wanted.
A: Construct the numbers by induction:
Let $a_n$ the $n$-digit number with digits $1$ or $2$ divisible by $2^n$.
If $\frac{a_n}{2^{n+1}}$ is an integer, then $a_{n+1}=2\cdot 10^n+a_n$ (prepend a $2$)
Otherwise,  $a_{n+1}=10^n+a_n$ (prepend a $1$)
It is easy to show that $a_{n+1}$ is divisible by $2^{n+1}$
It can be even shown that $a_n$ is the unique $n$ digit number of $1$'s and $2$'s that is divisible by $2^n$ (take two such numbers which obviously should be congruent modulo $2^n$, and prove the digits are identical also by induction, but starting with the last digit).
