2010 AIME I Problems/Problem 12
Contents
Problem
Let be an integer and let . Find the smallest value of such that for every partition of into two subsets, at least one of the subsets contains integers , , and (not necessarily distinct) such that .
Note: a partition of is a pair of sets , such that , .
Solution
We claim that is the minimal value of . Let the two partitioned sets be and ; we will try to partition and such that the condition is not satisfied. Without loss of generality, we place in . Then must be placed in , so must be placed in , and must be placed in . Then cannot be placed in any set, so we know is less than or equal to .
For , we can partition into and , and in neither set are there values where (since and and ). Thus .
Video Solution
~Shreyas S
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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