1989 AMC 12 第 25 题

Question

在一次由两支队伍参加的越野赛中，每队各有五名选手。获得第

n

名的选手为其队伍贡献

n

分，总分较低的队伍获胜。若所有选手的名次都不并列，那么可能出现多少种不同的获胜分数？

In a certain cross-country meet between two teams of five runners each, a runner who finishes in the

n

th position contributes

n

to his team’s score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?

10

13

27

120

126

答案：B

知识点：子集极端原理系统列举

难度评级：2270

小提示：

两支队伍的总分之和为

1+2+\cdots+10

The two team scores add to

1+2+\cdots+10

大提示：

确定获胜分数的最小值，并验证低于总分一半的每个整数都能取得

Determine the minimum winning score and verify that every integer below half the total is attainable

解答：

两队分数之和为

55

，所以获胜分数至多为

27

。其最小值为

1+2+3+4+5=15

。从

15

到

20

的每个分数都可由

\{1,2,3,4,k\}

取得，其中

k=5,\ldots,10

。从

21

到

25

的分数可由

\{1,2,3,k,10\}

取得，其中

k=5,\ldots,9

；而

26,27

分别可由

\{1,2,4,9,10\}

和

\{1,2,5,9,10\}

取得。因此，一共有

13

个可能的整数，即从

15

到

27

。

所以正确答案是 B。

The two scores sum to

55,

so the winning score is at most

27.

Its minimum is

1+2+3+4+5=15.

Every score from

15

through

20

is obtained by

\{1,2,3,4,k\}

for

k=5,\ldots,10.

Scores

21

through

25

use

\{1,2,3,k,10\}

for

k=5,\ldots,9,

and

26,27

use

\{1,2,4,9,10\}

and

\{1,2,5,9,10\}.

Thus all

13

integers from

15

through

27

are possible.

Thus the correct answer is B.

1989 AMC 12 第 25 题

答案：B

难度评级：2270

小提示：

大提示：

解答：

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