AMC8 1992

AMC8 1992 · Q19

AMC8 1992 · Q19. It mainly tests Pigeonhole principle.

The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?

州际高速公路上第5个出口和第26个出口之间的距离是118英里。如果任意两个出口相距至少5英里，那么在第5个和第26个出口之间，两个连续出口之间可能的最大英里数是多少？

(A) 8 8

(B) 13 13

(D) 47 47

(E) 98 98

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): There are 21 segments between the 5th and 26th exits. Using the minimum length of 5 miles, 20 segments would yield 20 × 5 = 100 miles. This leaves 118 − 100 = 18 miles for the other segment. OR There are 21 segments between the 5th and 26th exits. If each segment were its minimal 5 miles length, then the total distance between the 5th and 26th exits would be 105 miles. Since 118 − 105 = 13, all additional miles could occur between one pair of consecutive exits. Such a pair would be 5 + 13 = 18 miles apart.

答案 (C)：第5和第26出口间有21段路段。用最小长度5英里，20段为20 × 5 = 100英里。剩余118 − 100 = 18英里给另一段。或者第5和第26出口间有21段。若每段均为最小5英里，则总距离105英里。118 − 105 = 13，因此额外13英里可集中在相邻一对出口间，该对相距5 + 13 = 18英里。

Topics

CP.005 Pigeonhole principle