Difference between revisions of "2000 AIME I Problems/Problem 10"
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Now, substituting for <math>x_{50}</math>, we get <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}</math>, and the answer is <math>75+98=\boxed{173}</math>. | Now, substituting for <math>x_{50}</math>, we get <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}</math>, and the answer is <math>75+98=\boxed{173}</math>. | ||
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+ | == Solution 2 == | ||
+ | Consider <math>x_k</math> and <math>x_{k+1}</math>. Let <math>S</math> be the sum of the rest 98 terms. Then <math>x_k+k=S+x_{k+1}</math> and <math>x_{k+1}+(k+1)=S+x_k.</math> Eliminating <math>S</math> we have <math>x_{k+1}-x_k=-\dfrac{1}{2}.</math> So the sequence is arithmetic with common difference <math>-\dfrac{1}{2}.</math> | ||
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+ | In terms of <math>x_{50},</math> the sequence is <math>x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.</math> Therefore <math>x_{50}+50=99x_{50}-\dfrac{50}{2}</math>. We are done by solving for <math>x_{50}</math>. | ||
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+ | -JZ | ||
== See also == | == See also == |
Revision as of 15:40, 3 February 2020
Contents
Problem
A sequence of numbers has the property that, for every integer between and inclusive, the number is less than the sum of the other numbers. Given that where and are relatively prime positive integers, find .
Solution
Let the sum of all of the terms in the sequence be . Then for each integer , . Summing this up for all from ,
Now, substituting for , we get , and the answer is .
Solution 2
Consider and . Let be the sum of the rest 98 terms. Then and Eliminating we have So the sequence is arithmetic with common difference
In terms of the sequence is Therefore . We are done by solving for .
-JZ
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.