2011 AMC 12B Problem 17

First, $$\begin{aligned} h_1(x) &=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right) \\ &=\log_{10}\!\left(10^{10x-1}\right) \\ &=10x-1. \end{aligned}$$ Iterating, $h_n(x)=10^n x$ $-(1+10+\cdots+10^{n-1}).$ Therefore $h_n(1)$ is an $n$-digit integer whose units digit is $9$ and all of whose other digits are $8.$ For $n=2011,$ the digit sum is $$8\cdot2010+9=16{,}089.$$ Thus, the correct answer is B.