2011 AMC 12B Problems/Problem 17

$\text{Let }f(x)\text{ = }10^{10x}, g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right), h_{1}(x)\text{ = }g(f(x)),\text{and }h_{n}(x)\text{ = }h_{1}(h_{n-1}(x))\\\text{\\for integers }n\ge 2.\text{What is the sum of the digits of }h_{2011}(1)?$

$g(x)\text{ = }\text{log}_{10}\left(\frac{x}{10}\right)\text{ = }\text{log}_{10}\left({x}\right)\text{ - 1}$

$h_{1}(x)\text{ = }g(f(x))\text{ = }g(10^{10x})\text{ = }\text{log}_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$

Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ :

For n = 1, $h_{1}(x)\text{ = }10x - 1$

Assume $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n:

$h_{n+1}(x)\text{ = } h_{1}(h_{n}(x))\text{ = }10 h_{n}(x) - 1\text{ = 10 }(10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1 \\= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{n-1}) - 1$

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2011 AMC 12B Problems/Problem 17