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

$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$

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

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:

$\begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*}$

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 $\boxed{16089\textbf{(B)}}$

Solution 2 (Quick, Non-Rigorous Trends)

As before, $h_1(x)=10x-1$ . Compute $h_1$ , $h_2$ , and $h_3$ for $x=1$ to yield 9, 89, and 889. Notice how this trend will evidently repeat this trend (multiply by 10, subtract 1, repeat). As such, $h_2011$ is just 2010 8's followed by a nine. $2010(8)+9=\boxed{\textbf{B)}16089}$ .

~~BJHHar

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

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2011 AMC 12B Problems/Problem 17

Contents

Problem

Solution

Solution 2 (Quick, Non-Rigorous Trends)

See also