Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 12B Problems/Problem 17"
m (Added parentheses)
 
(5 intermediate revisions by 2 users not shown)
Line 8: Line 8:
 
<math>g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1</math>
 
<math>g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1</math>
  
−
<math>h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}</math>
+
<math>h_{1}(x)=g(f(x))\text{ = }g(10^{10x})=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}</math>
  
 
Proof by induction that <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math>:
 
Proof by induction that <math>h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})</math>:
Line 30: Line 30:
 
The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\textbf{(B)}}</math>
 
The sum of the digits is 8 times 2010 plus 9, or <math>\boxed{16089\textbf{(B)}}</math>
  
−
==Solution 2 (Non-Rigorous Trends)==
+
==Solution 2 (Quick, Non-Rigorous Trends)==
−
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(x)</math>, <math>h_2(x)</math>, and <math>h_3(x)</math> to yield 9, 89, and 889. Notice how this trend will 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}}

Latest revision as of 15:40, 21 September 2024

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 $h_1(x)$, $h_2(x)$, and $h_3(x)$ to yield 9, 89, and 889. Notice how this trend will 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_17&oldid=227728"