Difference between revisions of "2014 IMO Problems/Problem 1"
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Define <math>g(n) = a_0 + a_1 + \dots + a_n - n a_n</math>. Then because <math>a_{n+1} > a_n</math>, we have | Define <math>g(n) = a_0 + a_1 + \dots + a_n - n a_n</math>. Then because <math>a_{n+1} > a_n</math>, we have | ||
<cmath>a_0 + a_1 + \dots + a_n - n a_n > a_0 + a_1 + \dots + a_n + a_{n+1} - (n+1) a_{n+1}.</cmath> | <cmath>a_0 + a_1 + \dots + a_n - n a_n > a_0 + a_1 + \dots + a_n + a_{n+1} - (n+1) a_{n+1}.</cmath> | ||
− | Therefore, <math>g</math> is also monotonic decreasing. Note that <math>g(N+1) = a_0 + a_1 + \dots + a_{N+1} - (N+1) a_{N+1} | + | Therefore, <math>g</math> is also monotonic decreasing. Note that <math>g(N+1) = a_0 + a_1 + \dots + a_{N+1} - (N+1) a_{N+1} \le 0</math> from our inequality, and so <math>g(k) \le 0</math> for all <math>k > N</math>. Thus, the given inequality, which requires that <math>g(n) > 0</math>, cannot be satisfied for <math>n > N</math>, and so <math>N</math> is the unique solution to this inequality. |
--[[User:Suli|Suli]] 22:38, 7 February 2015 (EST) | --[[User:Suli|Suli]] 22:38, 7 February 2015 (EST) |
Latest revision as of 23:43, 7 February 2015
Contents
Problem
Let be an infinite sequence of positive integers, Prove that there exists a unique integer such that
Solution
Define . (In particular, ) Notice that because , we have Thus, ; i.e., is monotonic decreasing. Therefore, because , there exists a unique such that . In other words, This rearranges to give Define . Then because , we have Therefore, is also monotonic decreasing. Note that from our inequality, and so for all . Thus, the given inequality, which requires that , cannot be satisfied for , and so is the unique solution to this inequality.
--Suli 22:38, 7 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2014 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |