Maths Olympiad Prep

Library / /152 of 520

Combinatorics Difficulty 6.0 National olympiad Find the answer

8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of kk societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with 2m+12 m+1 students ( mm is a positive integer) is in exactly mm societies. Find all possible values of kk.

The above text is already in English. Here is the text with the same formatting and line breaks:

8. C1 (PUR) There are 10001 students at a university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of kk societies. Suppose that the following conditions hold: (i) Each pair of students are in exactly one club. (ii) For each student and each society, the student is in exactly one club of the society. (iii) Each club has an odd number of students. In addition, a club with 2m+12 m+1 students ( mm is a positive integer) is in exactly mm societies. Find all possible values of kk.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

8. Let us write n=10001 n = 10001 . Denote by T\mathcal{T} the set of ordered triples (a,C,S)(a, C, \mathcal{S}), where aa is a student, CC a club, and S\mathcal{S} a society such that aCa \in C and CSC \in \mathcal{S}. We shall count T|\mathcal{T}| in two different ways. Fix a student aa and a society S\mathcal{S}. By (ii), there is a unique club CC such that (a,C,S)T(a, C, \mathcal{S}) \in \mathcal{T}. Since the ordered pair (a,S)(a, \mathcal{S}) can be chosen in nkn k ways, we have that T=nk|\mathcal{T}| = n k. Now fix a club CC. By (iii), CC is in exactly (C1)/2(|C| - 1) / 2 societies, so there are C(C1)/2|C|(|C| - 1) / 2 triples from T\mathcal{T} with second coordinate CC. If C\mathcal{C} is the set of all clubs, we obtain T=CCC(C1)2|\mathcal{T}| = \sum_{C \in \mathcal{C}} \frac{|C|(|C| - 1)}{2}. But we also conclude from (i) that
CCC(C1)2=n(n1)2 \sum_{C \in \mathcal{C}} \frac{|C|(|C| - 1)}{2} = \frac{n(n - 1)}{2}
Therefore n(n1)2=nk\frac{n(n - 1)}{2} = n k, i.e., k=n12=5000k = \frac{n - 1}{2} = 5000. On the other hand, for k=n12k = \frac{n - 1}{2} there is a desired configuration with only one club CC that contains all students and kk identical societies with only one element (the club CC). It is easy to verify that (i)-(iii) hold.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.