2011 AMC 12B Problems/Problem 17

Problem

Let $f(x) = 10^{10x}, g(x) = \log_{10}\left(\frac{x}{10}\right), h_1(x) = g(f(x))$ , and $h_n(x) = h_1(h_{n-1}(x))$ for integers $n \geq 2$ . What is the sum of the digits of $h_{2011}(1)$ ?

$\textbf{(A)}\ 16081 \qquad \textbf{(B)}\ 16089 \qquad \textbf{(C)}\ 18089 \qquad \textbf{(D)}\ 18098 \qquad \textbf{(E)}\ 18099$

Solution

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

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

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

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

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

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

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\ \(\textbf{(B)}}$ (Error compiling LaTeX. Unknown error_msg)

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2011 AMC 12B Problems/Problem 17

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