Difference between revisions of "1984 USAMO Problems/Problem 4"
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Latest revision as of 12:35, 18 July 2016
Problem
A difficult mathematical competition consisted of a Part I and a Part II with a combined total of 
problems. Each contestant solved 
problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.
Solution
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See Also
| 1984 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
