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Difference between revisions of "2011 AMC 12B Problems/Problem 17"
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<math>h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})</math>, which is the 2011-digit number 8888...8889
 
<math>h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})</math>, which is the 2011-digit number 8888...8889
  
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The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\ \(\textbf{(B)}}</math>
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The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\textbf{(B)}}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}}
 
{{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:22, 16 May 2015

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

See also

2011 AMC 12B (Problems • Answer Key • Resources)
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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