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112 groups of 11 members each with one common in every two groups

Source: Greece JBMO TST 2015 p4

April 29, 2019
combinatoricsset theorySets

Problem Statement

Pupils of a school are divided into 112112 groups, of 1111 members each. Any two groups have exactly one common pupil. Prove that: a) there is a pupil that belongs to at least 1212 groups. b) there is a pupil that belongs to all the groups.