lets assume for $689$ : ,
$$689 = (6+ 8 + 9 ) = 23$$ $$23 = (2 + 3) = 5$$
We can get this way:
$$689 \!\!\mod \;9 = 5$$
I found a approach like this: $$10 = (9 * 1) + 1 $$ $$100 = (9 * 11) + 1 $$
this way : $$689 = 9 + (8*10) + (6*100)$$ $$689 = 9 + (8*(9+1)) + (6*(9 * 11 + 1))$$ $$689 = (9+8+6) + 9 * ((8 * 1) + (6 * 11))$$
But I couldn't figure out out how to proceed forward after this?
(Hint) :- Use modular arithmetic .
Note that $10^n \equiv 1\ (\textrm{mod}\ 9)$ for any positive integer $n$ and every number can be uniquely represented as $10^na_1 + 10^{n - 1}a_2 + ... + a_n$ where $a_1,a_2,...,a_n$ are positive integers from the set $[1,2,3,4,5,6,7,8,9]$ .
You have actually done the same thing in the case of $689$ . Since you got :- $$(9 + 8 + 6) + 9 * ((8*1) + (6*11))$$ We get :- $689 \equiv (9 + 8 + 6)\ (\textrm{mod}\ 9)$
=> $(9 + 8 + 6) \equiv 5\ (\textrm{mod}\ 9)$ , and that's the same dealing with the remainder of the sum of digits after division by $9$ .
Edit :- As mentioned by @Toby Mak, the remainder found from the division by $9$ is called it's digital root .