# Difference between revisions of "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 exponential 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$. Examining this relationship, we see that every new number$ (Error compiling LaTeX. )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}$). We can clearly see that there will be no solution k where k is even, as the$ (Error compiling LaTeX. )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$. Looking at the back of both sequences, we see that term k in$ (Error compiling LaTeX. )\{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$ (Error compiling LaTeX. )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, but I think this solution takes a different approach than the one above.

## Video Solution by Richard Rusczyk

~dolphin7

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