Yrurtj

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!

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 $1le nle 9$. Also, $1le d^*(n)le 9$, 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
$$
n=a_0+a_1cdot10+a_2cdot10^2+dots+a_ncdot10^n,
$$
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 $5$.

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

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 $3$.

The 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.