In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

A. 22

B. 23

C. 24

D. 26

E. 30

1. 2019 AMC 8 官方解析

在一个锦标赛中六支队伍每两队比赛两次. 胜者积3分，平局得1分，负者不得分. 当比赛都进行完最高得分的三队总分相同. 那么并列最高得分的三支队，最大可能的得分是？
A better way is to analyze the choice from largest to smallest. Could 30 be possible? Indeed there are total $6 \times 5 = 30$ games. If all are win-lose out comes,the total are 90 points. However 30 implies a team must have 10 wins. For the top three teams, they could' participate in games with both parties winning.What about 26? 26 has to be 8 wins and 2 draws. This is impossible: since no loses among these three, each team has to contribute 4 draws, twice with every other two.
24? 7 wins and 3 draws or 8 wins and 2 loses. Now it could be possible. For example they could form a cycle: each one wins twice to another and loses to the third one.If we need to be quite precise, consider why 25 , 27 , 28 don' work, those are longer.