Question

Mary、Alina、Tina 和 Hanna 四个女孩在音乐会上以三人组合唱歌，每首歌有一个女孩不唱。Hanna 唱了 $7$ 首，比任何其他女孩都多；Mary 唱了 $4$ 首，比任何其他女孩都少。这些三人组合一共唱了多少首歌？

Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?

$7$

$8$

$9$

$10$

$11$

答案：A

知识点：整除性双重计数极限情形界定

难度评级：1610

解答：

若共唱 $N$ 首歌，则总演唱人次为 $3N$ 。Alina 和 Tina 的演唱次数都严格介于 $4$ 和 $7$ 之间，所以各为 $5$ 或 $6$ 。

因此 $3N = 7 + 4 + (\text{Alina}) + (\text{Tina})$ ，可能为 $21, 22$ ，或 $23$ 。只有 $21$ 是 $3$ 的倍数，所以 $N = 7$ 。

所以正确答案是 A。

If $N$ songs are sung, the total number of girl-appearances is $3N.$ Alina and Tina each sang strictly between $4$ and $7,$ so each sang $5$ or $6.$

Then $3N = 7 + 4 + (\text{Alina}) + (\text{Tina}),$ which is $21, 22,$ or $23.$ Only $21$ is a multiple of $3,$ so $N = 7.$

Thus, the correct answer is A.

1999 AMC 12 第 14 题

答案：A

难度评级：1610

解答：

其他年份的第 14 题