The problem must be $ Q(3^k)\geq Q(3^{k + 1})$. In that case, the solution is rather well-known; suppose the statement doesn't happen. Then there exists some $ n_0$ such that for all integers $ n\geq n_0$, $ Q(3^n) < Q(3^{n + 1})$. But we all know that $ Q(k)\equiv k \pmod {9}$, so basically $ Q(3^n ) \equiv Q(3^{n + 1}) \pmod 9$, therefore $ Q(3^{n + 1})\geq 9 + Q(3^n)$, and by induction we get \[ Q(3^{n + k}) \geq 9k + Q(3^n), \] for all positive integers $ k$. Problem is that if $ k\to\infty$, then $ \lim_{k\to\infty} \frac { 3^{n + k} }{10^k} = 0$, so at some point $ k_0$ we have $ 3^{n + k_0} < 10^{k_0}$, which means that $ 3^{n + k_0}$ has at most $ k_0$ digits, so $ Q(3^{n + k_0})\leq 9\cdot k_0$, which is a contradiction.