Difference between revisions of "2012 AMC 12A Problems/Problem 24"
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Rearranging in decreasing order gives | 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>. | <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>. | ||
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+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://www.youtube.com/watch?v=f1nxu8MWWKc | ||
+ | |||
+ | ~dolphin7 | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:43, 4 April 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
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 .
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=f1nxu8MWWKc
~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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