How to prove that all integers with more than 1 digits, minus the sum of their digits result in $\mod 3=0$?

Question

I have ran a script on the first $100000$ natural numbers and noticed that for any Integer (with more than 1 digit) minus the sum of its digits will result in $\mod 3 = 0$

($w$ Starting Integer) → ($x$ sum of adding digits of $w$) → ($y = w - x$) → ($z = y/3$)

Examples:

$10 → 1 → 9 → 3$

$11 → 2 → 9 → 3$

$12 → 3 : 9 → 3$

$13 → 4 : 9 → 3$

$14 → 5 → 9 → 3$

$15 → 6 → 9 → 3$

$16 → 7 → 9 → 3$

$17 → 8 → 9 → 3$

$18 → 9 → 9 → 3$

$19 → 10 → 9 → 3$

$20 → 2 → 18 → 6$

$21 → 3 → 18 → 6$

$22 → 4 → 18 → 6$

I can see the logic in the heuristic, however I am struggling to propose a generalized proof for any number?