Difference between revisions of "2010 USAMO Problems/Problem 6"
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== Solution == | == Solution == | ||
− | + | To find this, assume only the two extremes, that is when the points have the same, with <math>nsub1</math> <math>-nsub1</math> and so on for 66 more times and then two zeroes. After I erase all 67 positive or negative for each one, I'm done. Henceforth we can get at least 67, because quite intuitively, we can erase more if this is not set up this way. | |
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== See Also == | == See Also == | ||
{{USAMO newbox|year=2010|num-b=5|after=Last question}} | {{USAMO newbox|year=2010|num-b=5|after=Last question}} |
Revision as of 16:34, 28 March 2012
Problem
A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer at most one of the pairs and is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point for each of the 68 pairs in which at least one integer is erased. Determine, with proof, the largest number of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.
Solution
To find this, assume only the two extremes, that is when the points have the same, with and so on for 66 more times and then two zeroes. After I erase all 67 positive or negative for each one, I'm done. Henceforth we can get at least 67, because quite intuitively, we can erase more if this is not set up this way.
See Also
2010 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |