$3^a\mid s(n) \Rightarrow 3^a\mid n$

Question

This is not a homework question, neither a championship problem (as far as I've searched in the net), and it came up noticing a singular pattern, involving the powers of $3$: "Prove or disprove that $$\forall a \in \Bbb{N},\qquad3^a\mid s(n) \Rightarrow 3^a\mid n$$ where $s(n)$ equals the sum of digits of a positive integer $n$ written in base $10$." This is obviously true for $a=0,1$ ($1\mid n\ \forall n \in \Bbb{N}$ and if $a=1$, it's simply the proof of the divisibility of an integer by $3$), but it seems valid also for $3^2,3^3,3^4,\ldots$ Any help is appreciated!

Answer 1

It's only true for $a \leqslant 2$. For $a = 3$, we have the counterexample $n = 1899$, with $s(n) = 27 = 3^3$, but $n = 3^2\cdot 211$. Generally,

$$s\left(9\cdot 10^{3^{a-3}} + (10^{3^{a-2}}-1)\right) = 3^a,$$

and $3^a \mid (10^{3^{a-2}}-1)$, but $3^a \nmid 9\cdot 10^{3^{a-3}}$ for $a > 2$.

Answer 2

$27$ divides $9+8+7+3$ but $27$ does not divide $9873$.