Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$

Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance $$ n=17254,\quad d(n)=1+7+2+5+4=19, \quad d(d(n))=1+9=10, \quad d(d(d(n)))=1+0=1=d^*(n) $$ Note that $d(n)\le n$, equality holding if and only if $1\le n\le 9$. Also, $1\le d^*(n)\le 9$, because the process stops only when the digit sum obtained is a one-digit number.

The main point is that $n-d(n)$ is divisible by $9$: indeed, if $$ n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n, $$ then $$ n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1) $$ and each factor $10^k-1$ is divisible by $9$. This extends to the reduced digit sum, because we can write (in the example above) $$ n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr) $$ and each parenthesized term is divisible by $9$. This works the same when a different number of steps is necessary.

Since the only one-digit number divisible by $9$ is $9$ itself, we can conclude that

$d^*(n)=9$ if and only if $n$ is divisible by $9$.

Since $360$ is divisible by $9$, its reduced digit sum is $9$; the same happenso for $180$ and so on.

When you divide an even integer $n$ by $2$, the quotient is divisible by $9$ if and only if $n$ is divisible by $9$.

What if the integer $n$ is odd? Well, the digits sum of $10n$ is the same as the digit sum of $n$. So what you are actually doing when arriving at $45$ is actually \begin{align} &45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\ &225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9 \end{align} and so on. Note that the first step can also be stated as $$ 45 \xrightarrow{\cdot5} 225 $$

Under these operations divisibility by $9$ is preserved, because you divide by $2$ or multiply by $5$.

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 $10^{a+b}$.

A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.