Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on Combinatorics in Tournament.
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two player earned \(\frac{1}{2}\) point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?
Integers
Combinatorics
Algebra
Answer: is 25.
AIME I, 1985, Question 14
Elementary Algebra by Hall and Knight
Let there be n+10 players
Case I from n players not in weakest 10, \({n \choose 2}\) games played and \({n \choose 2}\) points earned
Case II n players also earned \({n \choose 2}\) points against weakest 10
Case III now weakest 10 played among themselves \({10 \choose 2}\)=45 games and 45 points earned
Case IV 10 players also earned 45 points against stronger n
So total points earned= 2[\({n \choose 2}\)+45]=\(n^{2}-n+90\)
case V 1 point earned per game \({n+10 \choose 2}\)=\(\frac{(n+10)(n+9)}{2}\) games and \(\frac{(n+10)(n+9)}{2}\) points earned
So \(n^{2}-n+90=\frac{(n+10)(n+9)}{2}\)
or, \(n^{2}-21n+90=0\)
or, n=6, n=15 here taking n>10,
or, n=15 or, n+10=25.
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