Difference between revisions of "2000 AIME I Problems/Problem 10"
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Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus, <math>S = 2n + k</math>, so <math>n = \frac{S-k}{2}</math>. We then obtain the following sequence: | Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus, <math>S = 2n + k</math>, so <math>n = \frac{S-k}{2}</math>. We then obtain the following sequence: |
Revision as of 21:26, 29 December 2022
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, .
Solving, we get The answer is
- JZ
- edited by erinb28lms
Solution 3(very similar to 1)
Let be the value of , be the sum of the 99 other numbers, and be the sum of 100 numbers. Thus, , so . We then obtain the following sequence:
Solving for the left hand side, we get . This means . Since = , we get . Thus, the answer is . ~SoilMilk
Video solution
https://www.youtube.com/watch?v=TdvxgrSZTQw
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.