AMC8 1991

AMC8 1991 · Q14

AMC8 1991 · Q14. It mainly tests Casework, Extremal principle.

Several students are competing in a series of three races. A student earns 5 points for winning a race, 3 points for finishing second and 1 point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?

几名学生参加三场赛跑的系列赛。学生赢得比赛得5分，第二名得3分，第三名得1分。没有并列名次。一个学生在三场比赛中必须获得的最少积分，以保证比任何其他学生获得的积分都多，是多少？

(A) 9 9

(B) 10 10

(D) 13 13

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

If one student earns $5 + 5 + 5 = 15$ points, no other student can earn more than $3 + 3 + 3 = 9$ points. If one student earns $5 + 5 + 3 = 13$ points, no other student can earn more than $3 + 3 + 5 = 11$ points. However, if one student earns $5 + 3 + 3 = 11$ or $5 + 5 + 1 = 11$ points, some other student can earn $3 + 5 + 5 = 13$ or $3 + 3 + 5 = 11$ points. Thus 13 points is the smallest number of points a student must earn to be guaranteed of earning more points than any other student.

如果一个学生获得$5 + 5 + 5 = 15$分，没有其他学生能超过$3 + 3 + 3 = 9$分。如果一个学生获得$5 + 5 + 3 = 13$分，没有其他学生能超过$3 + 3 + 5 = 11$分。但是，如果一个学生获得$5 + 3 + 3 = 11$或$5 + 5 + 1 = 11$分，其他学生可能获得$3 + 5 + 5 = 13$或$3 + 3 + 5 = 11$分。因此13分是一个学生必须获得的最少积分，以保证比任何其他学生得分多。

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