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XXV OBM

Brazil counting and probability

Problem

Let be a set with elements. Take a positive integer . Let be any distinct subsets of . For each take or . Find the smallest such that we can always choose so that .
Solution
The sets , where or , are all disjoint. If , it follows that one of them must be empty. Hence its complement (which is , where ), is .

On the other hand if , then we can choose subsets such that all intersections are non-empty. That means all possible unions are incomplete. Thus the answer is the smallest such that .
Final answer
the smallest k such that 2^k > n (equivalently, floor(log2 n) + 1)

Techniques

Pigeonhole principle