2012 AMC 12A Problems/Problem 24

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 1

First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$ (decreasing exponential function), and $g(x) = x^k$ for $k > 0$ (increasing power function for positive $x$ ). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.

We will now examine 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(see Note 1).

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

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

We can use induction to prove this statement. (not necessary for 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 $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$ .

Solution 2

Start by looking at the first few terms and comparing them to each other. We can see that $a_1 < a_2$ , and that $a_1 < a_3 < a_2$ , and that $a_3 < a_4 < a_2$ , and that $a_3 < a_5 < a_4$ ...

From this, we find the pattern that $a_{k-1} < a_{k+1} < a_k$ .

Examining this relationship, we see that every new number $a_k$ will be between the previous two terms, $a_{k-1}$ and $a_{k-2}$ . Therefore, we can see that $a_1$ is the smallest number, $a_2$ is the largest number, and that all odd terms are less than even terms. Furthermore, we can see that for every odd k, $a_k < a_{k+2}$ , and for every even k, $a_k > a_{k+2}$

This means that rearranging the terms in descending order will first have all the even terms from $a_2$ to $a_{2012}$ , in that order, and then all odd terms from $a_{2011}$ to $a_1$ , in that order (so $\{b_k\}_{k=1}^{2011} = {a_2, a_4, a_6, ... a_{2008}, a_{2010}, a_{2011}, a_{2009}, ... a_5, a_3, a_1}$ ).

We can clearly see that there will be no solution k where k is even, as the $k$ th term in $\{a_k\}_{k=1}^{2011}$ will appear in the same position in its sequence as the $2k$ th term does in $\{a_k\}_{k=1}^{2011}$ , where k is even. Therefore, we only have to look at the odd terms of $a_k$ , which occur in the latter part of $b_k$ .

Looking at the back of both sequences, we see that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2012 - k$ , and that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2k - 1$ . Setting these two expressions equal to each other, we get $671$ .

However, remember that we started counting from the back of both sequences. So, plugging $671$ back into either side of the equation from earlier, we get our answer of $\boxed{\textbf{(C)} 1341}$ .

Sorry for the sloppy explanation. It's been two years since I've tried to give a solution to a problem, and this is the first time I've really used \LaTeX. But I think this solution takes a different approach than the one above.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/255

~dolphin7

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2012 AMC 12A Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

Video Solution by Richard Rusczyk

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