A logarithm problem

$\begin{aligned} h_1(x) & = g(f(x)) = g \left(10^{10x}\right) =\log_{10} \left(\frac {10^{10x}}{10} \right) = \log_{10} \left(10^{10x-1} \right) = 10x-1 \\ h_2(x) & = h_1(h_1(x)) = 10(10x-1)-1 = 10^2x - 11 = 10^2x - \frac {10^2-1}9 \\ h_3(x) & = 10(10^2x-11)-1 = 10^3x - 111 = 10^3x - \frac {10^3-1}9 \end{aligned}$

$\begin{aligned} \implies h_n (x) & = 10^nx - \frac {10^n - 1}9 & \small \color{#3D99F6} \text{See proof by induction below.} \\ h_{2011} (1) & = 10^{2011} - \frac {10^{2011}-1}9 \\ & = 1\underbrace{000000\cdots 000000}_{\text{Number of 0's = 2011}} - \underbrace{111111\cdots 111111}_{\text{Number of 1's = 2011}} \\ & = \underbrace{888888\cdots 888888}_{\text{Number of 8's = 2010}}9 \end{aligned}$

The sum of digits of $h_{2011}(1)$ is $2010\times 8 + 9 = 16089 = 173 \times 93$ . Therefore $Q=\boxed{93}$ .

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Proof by induction: The claim $h_n (x) = 10^n x - \dfrac {10^n-1}9$ . When $n=1$ , $h_1(x) = 10^1x - \dfrac {10-1}9 = 10x - 1$ as given. Therefore the claim is true for $n=1$ . Now assuming that it is true for $n$ , then:

$\begin{aligned} h_{n+1} & = h_1(h_n(x)) = 10 \left(10^n x - \frac {10^n-1}9\right) - 1 = 10^{n+1}x - \frac {10^{n+1} - 10 + 9}9 = 10^{n+1}x - \frac {10^{n+1} - 1}9 \end{aligned}$

Therefore, the claim is true for $n+1$ and hence true for all $n\ge 1$ .

$h_1(x)=\log_{10}\left(\dfrac{10^{10x}}{10}\right)=10x-1$ ,so $h_1(1)=9,h_2(1)=89,h_3(1)=889,...,h_n(1)=\underbrace{888 \cdots 8889}_{\text{Number of 8s }= n-1}$

The digit sum of $h_{2011}(1)$ is $8\times2011+1=173\times\boxed{93}$

We are given $f\left(x\right)=10^{10x}$ , $g\left(x\right)=\log\left(\frac{x}{10}\right)$ , $h_1(x) = g(f(x)) = \log(10^{10x - 1}) = 10x - 1$ , and $h_n(x) = h_1(h_{n-1}(x))$ .

Looking at the first couple functions in the sequence $h_n(x)$ :

• $h_1(x) = 10x - 1$

• $h_2(x) = 10(10x - 1) - 1 = 100x - 11$

• $h_3(x) = 10(100x - 11) - 1 = 1000x - 111$

It is clear that $h_n(x) = 10^nx - \frac{10^n - 1}{9}$ . If $x = 1$ , then $h_n(1) = 888...9$ for $n - 1$ occurrences of 8.

The digit sum of $h_n(1)$ is thereby given by $s(n) = 9 + 8(n - 1)$ . $s(2011) = 9 + 8 * 2010 = 16089 = 173 * 93$ . Therefore, $\boxed{Q = 93}$ .