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Difference between revisions of "2011 AMC 12B Problems/Problem 17"
(→‎Solution 2 (Non-Rigorous Trends))
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==Solution 2 (Non-Rigorous Trends)==
 
==Solution 2 (Non-Rigorous Trends)==
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As before, combine the functions to see that <math>h_1(x)=10x-1</math>. Compute <math>h_1</math>, <math>h_2</math>, and <math>h_3</math> for <math>x=1</math> to yield 9, 89, and 889. In computing, notice how this trend will evidently continue, because it repeats (multiply by 10 and subtract 1. Multiply by 10 and subtract 1. And so forth). As such, <math>h_2011</math> is 2010 8's followed by a nine. <math>2010(8)+9=\boxed{\textbf{B)}16089}</math>.
+
As before, <math>h_1(x)=10x-1</math>. Compute <math>h_1</math>, <math>h_2</math>, and <math>h_3</math> for <math>x=1</math> to yield 9, 89, and 889. Notice how this trend will evidently repeat this trend (multiply by 10, subtract 1, repeat). As such, <math>h_2011</math> is just 2010 8's followed by a nine. <math>2010(8)+9=\boxed{\textbf{B)}16089}</math>.
  
 
~~BJHHar
 
~~BJHHar
−
 
  
 
== 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 19:37, 27 January 2020

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 (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

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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