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

Revision as of 12:07, 15 May 2020

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

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

Video Solution by Richard Rusczyk

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

~dolphin7

@@ Line 50: / Line 50: @@
 ==Video Solution by Richard Rusczyk==
-https://www.youtube.com/watch?v=f1nxu8MWWKc
+https://artofproblemsolving.com/videos/amc/2012amc12a/255
 ~dolphin7
 ==See Also==
 {{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}}
 {{MAA Notice}}

2012 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2012 AMC 12A Problems/Problem 24"

Revision as of 12:07, 15 May 2020

Contents

Problem

Solution

Video Solution by Richard Rusczyk

See Also