Problem:
Find, with proof, the least integer such that if any 2016 elements are removed from the set , one can still find 2016 distinct numbers among the remaining elements with sum .
Solution:
The answer is
To see that must be at least this large, simply consider the situation when , 2016 are removed. Then among the remaining elements, any sum of 2016 elements is certainly at least .
Now we show this value of works. Consider the 3024 pairs of numbers , . After the elements of are deleted, at least of these pairs have both elements remaining. Since each pair has sum 6049, we can take these pairs to be the desired numbers.
The problems on this page are the property of the MAA's American Mathematics Competitions