AMC8 1990

AMC8 1990 · Q19

AMC8 1990 · Q19. It mainly tests Pigeonhole principle, Casework.

There are 120 seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

一排有 120 个座位。必须占用多少个座位，才能使下一个坐下的人必须坐在某人旁边？（最少）

(A) 30 30

(B) 40 40

(D) 60 60

(E) 119 119

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): In order for the fewest number of seats to be occupied, there must be someone in every third seat, beginning with \#2 and ending with \#119. There are a total of $\frac{120}{3} = 40$ occupied seats. OR Consider some simpler cases and make a table: Number of seats in the row: \[3 \quad 6 \quad 9 \quad 12 \] Number of occupied seats in the row: \[1 \quad 2 \quad 3 \quad 4\] In each case, the middle seat in every group of three seats must be occupied, so the desired number of occupied seats in a row of 120 seats is $\frac{120}{3} = 40$.

答案（B）：为了使被占用的座位数最少，必须每隔两个座位坐一个人，也就是从 \#2 开始到 \#119 结束。被占用的座位总数为 $\frac{120}{3} = 40$。或者考虑一些更简单的情况，并列出表格：一排座位数：\[3 \quad 6 \quad 9 \quad 12 \] 被占用的座位数：\[1 \quad 2 \quad 3 \quad 4\] 在每种情况下，每三个座位为一组时，中间的座位必须被占用，因此一排 120 个座位时，被占用的座位数为 $\frac{120}{3} = 40$。

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