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

(Added another solution (very sloppy but I think it's different))
(Added another solution (very sloppy but I think it's different))
Line 52: Line 52:
 
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> ...  
 
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 < </math>a_k+1 < <math>a_k</math>.
+
From this, we find the pattern that <math>a_k-1 < a_k+1 < </math>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>
+
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>).
+
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>.
+
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>.
+
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>.
+
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}$.
  
 
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.  
 
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.  

Revision as of 22:28, 30 December 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 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. ! Missing $ inserted.)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. ! Missing $ inserted.)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. ! Missing $ inserted.)\{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. ! Missing $ inserted.)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

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

~dolphin7

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS