在一场循环赛中，一共有6支队伍，每两支队伍之间都要比赛2次。某个队伍赢一次得3分，平局得1分，输了则得0分。当所有比赛都结束后，结果发现排名前三的队伍得分相同。问排名前三的队伍每支队伍最多可能得多少分？

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.