Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
\begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*}
As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!

1)  The sum of the digits of a multiple of $9$ will be a multiple of $9$.
Why?  If  $N$ is a multiple of $9$ then $N - 9k$ will also be a multiple of $9$ for all integers $k$.
If  $N = \sum_{k=0}^n a_k 10^k$ is a multiple of $9$ then $(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$ is a multiple of $9$.
And  $(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$
$(\sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$
$\sum_{k=0}^n [a_k*10^k - \underbrace{999...9}k*a_k]=$
$\sum_{k=0}^n a_k(10^k - \underbrace{999...9}k)=\sum_{k=0}^n a_k*1 =$ the sum of the digits of $N$.
2) we can extend that decimals:
If $N$ is $9$ times some terminating decimal than the sum of the digits are a multiple of $9$.
Why?  Well, its the same as above except $N$ has be divided by a power of $10$.  But that only affects the position of the decimal point.  It doesn't affect the digits. 
3) If $N$ is a $9$ times a terminating decimal than $\frac N2$ is $9$ times a terminating decimal.
If $N = 9\times M$ and $M$ is a terminating decimal then $\frac M2$ is a terminating decimal.  Then $\frac N2 = 9\times \frac M2$ is also $9$ times a terminating decimal and the digits add up to a multiple of $9$.
4) Since $360 = 9*40$ we start with a number whose digits add to $9$.  Successively dividing by $2$ will result in $9*20$ and $9*10$ and so on, always $9$ times a terminating decimal with the sum of the digits being a multiple of $9$.
