2012 AMC 12A Problems/Problem 24
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. ! Missing $ inserted.)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. ! Missing $ inserted.)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. ! Missing $ inserted.)\{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. ! Missing $ inserted.)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 |
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