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!

In general if an integer is divisible by 9 then its digital root is $9$.  Any multiple of an integer divisible by $9$ will also be divisible by $9$ and have digital root $9$.
What's slightly surprising here is that this property is also preserved by division by $2$.  Note that it doesn't work for, say, division by $3$, since $360/3 = 120$ has digital root $3$.
A relevant property that $2$ has, but $3$ doesn't, is that it's a factor of $10$. Dividing by $2$ is equivalent to multiplying by $5$, which preserves the property, and then dividing by $10$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.
More generally, division by an integer $n$ will preserve the property of having digital root $9$ if all the prime factors of $n$ are factors of $10$, that is, it equals $2^a5^b$ for some $a,b \geq1$.  This works because division by $2^a5^b$ is equivalent to multiplying by $2^b5^a$ and then dividing by $10^{a+b}$.  
