A little observation.

Question

solomonzelman · Answer

$\color{blue}{x=10b+a}$
$\color{blue}{y=10a+b}$

A difference between $\color{blue}{x}$ and $\color{blue}{y}$, is always a factor of $\color{blue}{9}$, unless it is $\color{blue}{0}$.

$\color{blue}{D(x,y):=|x-y|=9|a-b|}$.

So the difference between any two-digit number, and the same number but with a difference order of digits, is divisible by 9 (or is 0).

We know that the difference $\color{blue}{D}$ is a factor of 9, or 0. That means that the sum of digits is divisible by 9 (or the difference is 0).

When you add $n$ number of digits to $x$ (in any order), and $n$ number of digits to $y$ (in any order), as long as these digits are the same, the sum of digits of $\color{blue}{D}$ doesn't change.

This makes a conclusion:
$\color{black}{\rm The~~difference~~between~~any~~number~~and~~the~~same~~number}$
$\color{black}{\rm with~~digits~~rearranged~~is~~always~~divisible~~by~~9~~(or~~it~~is~~0).}$

solomonzelman · Answer

I had a few typos, but I hope it's readable.
Also:  10a+b, and 10b+a  is just a mathematical way that I chose to record numbers with digits ab, and digits ba (respectively).

faiqraees · Answer

$$\large\rm Proof: $$Let us consider a number which can be rewritten  in the form $a*10^n+b*10^{n-1}+c*10^{n-2}....$ 

Recall that$\large\rm 10≡1\mod9 $

Now we can rewrite the above number in the form
$\large\rm a*10^n+b*10^{n-1}+c*10^{n-2}....≡a*1+b*1+c*1.... \mod9$ 

We can conclude that if the digits when added up is divisible by 9, the number itself is also divisible by 9