2011 AMC 12B Problems/Problem 17

Revision as of 11:31, 11 October 2018 by Copeland (talk | contribs) (Solution)

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:

\[hn+1(x)=h1(hn(x))=10hn(x)1=10(10nx(1+10+102+...+10n1))1=10n+1x(10+102+...+10n)1=10n+1x(1+10+102+...+10(n+1)1)\]

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

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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