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

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

Problem

Solution

Solution 1