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)) |
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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 < | + | 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< | + | 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< | + | 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< | + | 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}< | + | 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< | + | 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}$. |
Sorry for the sloppy explanation. It's been two years since I've tried to give a solution to a problem, but I think this solution takes a different approach than the one above. | Sorry for the sloppy explanation. It's been two years since I've tried to give a solution to a problem, but I think this solution takes a different approach than the one above. |
Revision as of 23:28, 30 December 2020
Problem
Let be the sequence of real numbers defined by , and in general,
Rearranging the numbers in the sequence in decreasing order produces a new sequence . What is the sum of all integers , , such that
Solution 1
First, we must understand two important functions: for (decreasing exponential function), and for (increasing exponential function for positive ). is used to establish inequalities when we change the exponent and keep the base constant. is used to establish inequalities when we change the base and keep the exponent constant.
We will now examine the first few terms.
Comparing and , .
Therefore, .
Comparing and , .
Comparing and , .
Therefore, .
Comparing and , .
Comparing and , .
Therefore, .
Continuing in this manner, it is easy to see a pattern(see Note 1).
Therefore, the only when is when . Solving gives .
Note 1:
We claim that .
We can use induction to prove this statement. (not necessary for AMC):
Base Case: We have already shown the base case above, where .
Inductive Step:
Rearranging in decreasing order gives .
Solution 2
Start by looking at the first few terms and comparing them to each other. We can see that , and that , and that , and that ...
From this, we find the pattern that a_k$.
Examining this relationship, we see that every new number$ (Error compiling LaTeX. )a_ka_k-1a_k-2a_1a_2a_k < a_k+2a_k > a_k+2a_2a_2012a_2011a_1\{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\{a_k\}_{k=1}^{2011}2k\{a_k\}_{k=1}^{2011}a_kb_k$.
Looking at the back of both sequences, we see that term k in$ (Error compiling LaTeX. )\{a_k\}_{k=1}^{2011}2012 - k\{a_k\}_{k=1}^{2011}2k - 1671$.
However, remember that we started counting from the back of both sequences. So, plugging$ (Error compiling LaTeX. )671\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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.