# The Minimum Number Of Subsets

Find the last three digits of the smallest positive integer $$n$$ with the following property:

Let $$\mathcal{S}$$ be any set containing $$n$$ elements. Partition the $$5$$ element subsets of $$\mathcal{S}$$ into two partitions. Then, at least one of the partitions must contain $$2014$$ pairwise disjoint sets.

Details and assumptions

• Sets $$A_1, A_2, \cdots , A_k$$ are called pairwise disjoint if $$|A_i \cap A_j| = 0$$ for all $$i \neq j.$$

• This condition must hold for all sets $$\mathcal{S}$$ containing $$n$$ elements and all partitions of its $$5$$ element subsets.

• This problem is a generalization of an old USAMO problem.

• If the last three digits of $$n$$ are $$012,$$ enter $$12$$ as your answer.

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