Difference between revisions of "2012 AMC 12A Problems/Problem 24"
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
+ | We begin our solution by understanding two important functions: <math>f(x) = b^x</math> for <math>0 < b < 1</math>, and <math>g(x) = x^k</math> for <math>k > 0</math>. The first function is a decreasing exponential function. This means that for numbers <math>m > n</math>, <math>f(m) < f(n)</math>. The second function is an increasing function on the interval <math>[0, \infty]</math>. This means that for numbers <math>m > n</math>, <math>g(m) > g(n)</math>. <math>f(x)</math> is used to establish inequalities when we change the exponent keep the base constant. <math>g(x)</math> is used to establish inequalities when we change the base and keep the exponent constant. | ||
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+ | We will now begin by examining 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>. | ||
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+ | 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>. | ||
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+ | 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 < 0</math>. | ||
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+ | Therefore, <math>0 < a_1 < a_3 < a_2 < 1</math>. | ||
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+ | Continuing in this manner, it is easy to see a pattern. | ||
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+ | We claim that <math>0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1</math>. | ||
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+ | We will now use induction to prove this statement. (Note that this is not necessary on the AMC): | ||
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+ | 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>. | ||
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+ | 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>. | ||
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{{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2012|ab=A|num-b=23|num-a=25}} |
Revision as of 21:46, 4 August 2012
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
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\2012$ (Error compiling LaTeX. )
Solution
Solution 1
We begin our solution by understanding two important functions: for , and for . The first function is a decreasing exponential function. This means that for numbers , . The second function is an increasing function on the interval . This means that for numbers , . is used to establish inequalities when we change the exponent keep the base constant. is used to establish inequalities when we change the base and keep the exponent constant.
We will now begin by examining the first few terms.
Comparing and , .
Comparing and , .
Comparing and , .
Therefore, .
Continuing in this manner, it is easy to see a pattern.
We claim that .
We will now use induction to prove this statement. (Note that this is not necessary on the AMC):
Rearranging in decreasing order gives .
Therefore, the only when is when . Solving gives .
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |