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

(Created page with "== Problem == Let <math>\{a_k\}_{k=1}^{2011}</math> be the sequence of real numbers defined by <math>a_1=0.201,</math> <math>a_2=(0.2011)^{a_1},</math> <math>a_3=(0.20101)^{a_2}...")
 
(→‎See Also)
 
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Rearranging the numbers in the sequence  <math>\{a_k\}_{k=1}^{2011}</math> in decreasing order produces a new sequence  <math>\{b_k\}_{k=1}^{2011}</math>.  What is the sum of all integers <math>k</math>, <math>1\le k \le 2011</math>, such that <math>a_k=b_k?</math>
 
Rearranging the numbers in the sequence  <math>\{a_k\}_{k=1}^{2011}</math> in decreasing order produces a new sequence  <math>\{b_k\}_{k=1}^{2011}</math>.  What is the sum of all integers <math>k</math>, <math>1\le k \le 2011</math>, such that <math>a_k=b_k?</math>
  
−
<math> \textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\2012 </math>
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<math> \textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012 </math>
  
 +
== Solution 1 ==
  
−
== Solution ==
+
First, we must understand two important functions: <math>f(x) = b^x</math> for <math>0 < b < 1</math>(decreasing exponential function), and <math>g(x) = x^k</math> for <math>k > 0</math>(increasing power function for positive <math>x</math>). <math>f(x)</math> is used to establish inequalities when we change the ''exponent'' and keep the ''base'' constant. <math>g(x)</math> is used to establish inequalities when we change the ''base'' and keep the ''exponent'' constant.
 +
 
 +
We will now examine the first few terms.
 +
 
 +
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>.
 +
 
 +
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_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>.
 +
 
 +
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>.
 +
 
 +
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>.
 +
 
 +
Therefore, <math>0 < a_1 < a_3 < a_4 < a_2 < 1</math>.
 +
 
 +
Continuing in this manner, it is easy to see a pattern(see Note 1).
 +
 
 +
Therefore, the only <math>k</math> when <math>a_k = b_k</math> is when <math>2(k-1006) = 2011 - k</math>. Solving gives <math>\boxed{\textbf{(C)} 1341}</math>.
 +
 
 +
 
 +
'''Note 1''':
 +
We claim that <math>0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1</math>.
 +
 
 +
We can use induction to prove this statement. (not necessary for AMC):
 +
 
 +
Base Case: We have already shown the base case above, where <math>0 < a_1 < a_2 < 1</math>.
 +
 
 +
Inductive Step:
 +
 
 +
Rearranging in decreasing order gives
 +
<math>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</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Start by looking at the first few terms and comparing them to each other. We can see that <math>a_1 < a_2</math>, and that <math>a_1 < a_3 < a_2</math>, and that <math>a_3 < a_4 < a_2</math>, and that <math>a_3 < a_5 < a_4</math> ...
 +
 
 +
From this, we find the pattern that <math>a_{k-1} < a_{k+1} < a_k</math>.
 +
 
 +
Examining this relationship, we see that every new number <math>a_k</math> will be between the previous two terms, <math>a_{k-1}</math> and <math>a_{k-2}</math>. Therefore, we can see that <math>a_1</math> is the smallest number, <math>a_2</math> is the largest number, and that all odd terms are less than even terms. Furthermore, we can see that for every odd k, <math>a_k < a_{k+2}</math>, and for every even k, <math>a_k > a_{k+2}</math>
 +
 
 +
This means that rearranging the terms in descending order will first have all the even terms from <math>a_2</math> to <math>a_{2012}</math>, in that order, and then all odd terms from <math>a_{2011}</math> to <math>a_1</math>, in that order (so <math>\{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}</math>).
 +
 
 +
We can clearly see that there will be no solution k where k is even, as the <math>k</math>th term in <math>\{a_k\}_{k=1}^{2011}</math> will appear in the same position in its sequence as the <math>2k</math>th term does in <math>\{a_k\}_{k=1}^{2011}</math>, where k is even. Therefore, we only have to look at the odd terms of <math>a_k</math>, which occur in the latter part of <math>b_k</math>.
 +
 
 +
Looking at the back of both sequences, we see that term k in <math>\{a_k\}_{k=1}^{2011}</math> progresses backwards in the equation <math>2012 - k</math>, and that term k in <math>\{a_k\}_{k=1}^{2011}</math> progresses backwards in the equation <math>2k - 1</math>. Setting these two expressions equal to each other, we get <math>671</math>.
 +
 
 +
However, remember that we started counting from the back of both sequences. So, plugging <math>671</math> back into either side of the equation from earlier, we get our answer of <math>\boxed{\textbf{(C)} 1341}</math>.
 +
 
 +
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
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}}
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:38, 31 March 2024

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

See Also

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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