smc

$g(x)={\log }_{10}\left(\frac{x}{10}\right)={\log }_{10}\left(x\right)-1$ $h_1(x)=g(f(x))\text{ = }g(10^{10x}={\log }_{10}\left(10^{10x}\right)-1=10x-1$ Proof by induction that $h_n(x)\text{ = }10^{n}x-(1+10+10^2+...+10^{n-1})$: For $n=1$, $h_1(x)=10x-1$ Assume $h_n(x)=10^{n}x-(1+10+10^2+...+10^{n-1})$ is true for n: $$\begin{array}{rl}{h}_{n+1}(x)& =h_1(h_n(x))\\ & =10h_n(x)-1\\ & =10(10^{n}x-(1+10+10^2+...+10^{n-1}))-1\\ & =10^{n+1}x-(10+10^2+...+10^n)-1\\ & =10^{n+1}x-(1+10+10^2+...+10^{(n+1)-1})\end{array}$$ Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n. $h_{2011}(1)=10^{2011}\times 1-(1+10+10^2+...+10^{2010})$, which is the 2011-digit number 8888...8889 The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089}$