Prove that the digits in $\frac{n}{2^k}$ (where $n$ is an integer divisible by 9 and $k$ is any integer) sum to a multiple of 9? I don't know where to start.  Here are two examples of the statement:
$36/8=4.5$ and  $(4+5)\mod9=0$
$9/4=2.25$ and $(2+2+5)\mod9=0$

Hint $\ $ Multiply by a power of $10$ to get an integer multiple of $9$ with the same digits.
$\qquad x= \dfrac{9j}{2^k}\ \Rightarrow\ 10^k x = 5^k(9j)\ $  has same digits as $\,x\, $ (shifted left by $\,k$), except $0$'s at end
E.g. in your case $\ x = \dfrac{9}{2^2} = \color{#c00}{2.25}\ $ so $\,10^2 x = 5^2\cdot 9 = \color{#c00}{225}.\,$ 
and the other  is  $\,\ x = \dfrac{36}{2^3} = \color{#c00}{4.5}\ $ so $\,10^3 x = 5^3\cdot 36 = \color{#c00}{4500}$.
Being a multiple of $9$ we know by casting out nines that $9$ divides its digit sum (which is not altered by any trailing $0$'s).
