1999 AMC 12 第 9 题

先试着解答 1999 AMC 12 第 9 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1999 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

9.

Ashley 开始三小时车程前,汽车里程表读数为 2979229792,这是一个回文数,也就是从左到右和从右到左读都相同。到达目的地时,里程表又显示另一个回文数。若 Ashley 从未超过每小时 7575 英里的限速,下列哪个是她最大的可能平均速度?
Before Ashley started a three-hour drive, her car’s odometer reading was 29792,29792, a palindrome. (A palindrome is a number that reads the same way from left to right as it does from right to left.) At her destination, the odometer reading was another palindrome. If Ashley never exceeded the speed limit of 7575 miles per hour, which of the following was her greatest possible average speed?
331333\tfrac13
531353\tfrac13
662366\tfrac23
701370\tfrac13
741374\tfrac13
答案:D
知识点:回文数路程、速度与时间极限情形界定
小提示:
列出 2979229792 之后的几个回文数。
List the palindromes just above 2979229792
大提示:
在 33 小时内,行驶距离不能超过 3⋅75=2253 \cdot 75 = 225 英里。
The distance in 33 hours cannot exceed 3⋅75=2253 \cdot 75 = 225 miles
解答:
2979229792 之后的回文数依次为 29892,29992,3000329892, 29992, 30003 和 3010330103。三小时内最多行驶 3⋅75=2253 \cdot 75 = 225 英里。
到 3010330103 需要行驶 30103−29792=31130103 - 29792 = 311 英里,太远。到 3000330003 需要行驶 30003−29792=21130003 - 29792 = 211 英里,平均速度为 2113=7013\dfrac{211}{3} = 70\tfrac13 英里每小时。
所以正确答案是 D。
The palindromes after 2979229792 are 29892,29992,30003,29892, 29992, 30003, and 30103.30103. In three hours Ashley can drive at most 3⋅75=2253 \cdot 75 = 225 miles.
Reaching 3010330103 would require 30103−29792=31130103 - 29792 = 311 miles, which is too far. Reaching 3000330003 requires 30003−29792=21130003 - 29792 = 211 miles, giving average speed 2113=7013\dfrac{211}{3} = 70\tfrac13 miles per hour.
Thus, the correct answer is D.
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