Difference between revisions of "2012 AMC 12A Problems/Problem 24"

(Problem)
(Solution 1)
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Comparing <math>a_1</math> and <math>a_2</math>, <math>0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1</math>.
 
Comparing <math>a_1</math> and <math>a_2</math>, <math>0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1</math>.
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 +
Therefore, <math>0 < a_1 < a_2 < 1</math>.
  
 
Comparing <math>a_2</math> and <math>a_3</math>, <math>0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1</math>.
 
Comparing <math>a_2</math> and <math>a_3</math>, <math>0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1</math>.
  
Comparing <math>a_1</math> and <math>a_3</math>, <math>0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 0</math>.
+
Comparing <math>a_1</math> and <math>a_3</math>, <math>0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1</math>.
  
 
Therefore, <math>0 < a_1 < a_3 < a_2 < 1</math>.
 
Therefore, <math>0 < a_1 < a_3 < a_2 < 1</math>.
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 +
Comparing <math>a_3</math> and <math>a_4</math>, <math>0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1</math>.
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Comparing <math>a_2</math> and <math>a_4</math>, <math>0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1</math>.
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 +
Therefore, <math>0 < a_1 < a_3 < a_4 < a_2 < 1</math>.
  
 
Continuing in this manner, it is easy to see a pattern.
 
Continuing in this manner, it is easy to see a pattern.
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We will now use induction to prove this statement. (Note that this is not necessary on the AMC):
 
We will now use induction to prove this statement. (Note that this is not necessary on the AMC):
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 +
Base Case: We have already shown the base case above, where <math>0 < a_1 < a_2 < 1</math>.
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Inductive Step:
  
  

Revision as of 15:35, 5 August 2012

Problem

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general,

\[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\]

Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$

$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$

Solution

Solution 1

We begin our solution by understanding two important functions: $f(x) = b^x$ for $0 < b < 1$, and $g(x) = x^k$ for $k > 0$. The first function is a decreasing exponential function. This means that for numbers $m > n$, $f(m) < f(n)$. The second function is an increasing function on the interval $[0, \infty]$. This means that for numbers $m > n$, $g(m) > g(n)$. $f(x)$ is used to establish inequalities when we change the exponent keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.

We will now begin by examining the first few terms.

Comparing $a_1$ and $a_2$, $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$.

Therefore, $0 < a_1 < a_2 < 1$.

Comparing $a_2$ and $a_3$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$.

Comparing $a_1$ and $a_3$, $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$.

Therefore, $0 < a_1 < a_3 < a_2 < 1$.

Comparing $a_3$ and $a_4$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$.

Comparing $a_2$ and $a_4$, $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$.

Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$.

Continuing in this manner, it is easy to see a pattern.

We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$.

We will now use induction to prove this statement. (Note that this is not necessary on the AMC):

Base Case: We have already shown the base case above, where $0 < a_1 < a_2 < 1$.

Inductive Step:


Rearranging in decreasing order gives

$1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} =  a_1 > 0$.

Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{\textbf{(C)} 1341}$.


2012 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 12 Problems and Solutions
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