Olympiad Maths Prep

Library / /13 of 33

Combinatorics Difficulty 5.9 AIME, harder Prove it Belarus

Students of two groups decided to organize a chess tournament where each student from the first group plays exactly one game with each student from the other group. But one student from the first group and one student from the other one, due to some reasons, failed to participate in the tournament, so the total number of the games in the tournament has been 20% smaller than that of the games planned.
Find all possible numbers of the students participated in the tournament.

Solution

Answer: 17, 20, 29.
Let nn and mm be the numbers of students of the first, and respectively the second group that were initially supposed to participate in the tournament. Then S=(n−1)+(m−1)S = (n-1)+(m-1) students have taken part in the tournament. By condition,
(n−1)(m−1)=0.8 nm  ⟺  (m−5)(n−5)=20. (n-1)(m-1) = 0.8 \text{ nm} \iff (m-5)(n-5) = 20.
Therefore, the pair (n−5,m−5)(n-5, m-5) is one of three pairs: (1, 20), (2, 10), (4, 5). Thus, (n−5)+(m−5)(n-5) + (m-5) is equal to 1+20=211+20=21, or 2+10=122+10=12, or 4+5=94+5=9. Thus m+nm+n is equal to 31, or 22, or 19.
It is easy to see that all these pairs (m,n)(m, n) satisfy the problem conditions.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.