Divisibility of $abc-(a+b+c)$

We have $[100<n<1000; n\in\mathbb N]$ $n$ is three digit which can be expressed as $n=100x+10y+z$ were $0≤(x,y,z)<10$ . $\small\begin{aligned}\text{difference (d)}=n-(x+y+z)=(100x+10y+z)-(x+y+z)=99x+9y; d\Rightarrow 3(33x+3y)\Rightarrow 9(11x+y)\end{aligned}$ Therefore, $d\equiv0\mod(3)\equiv0\mod(9)$ is always true .