1994 AMC 8 第 21 题

先试着解答 1994 AMC 8 第 21 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 1994 AMC 8 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。

21.

一台口香糖球机中有 99 个红色、77 个白色和 88 个蓝色口香糖球。为了保证得到四个同色口香糖球,一个人至少必须购买多少个口香糖球?
A gumball machine contains 99 red, 77 white, and 88 blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
88
99
1010
1212
1818
答案:C
知识点:抽屉原理
小提示:
考虑最坏情况:最多能抽到多少个球而每种颜色都不超过三个?
Consider the worst case: how many gumballs can you draw with at most three of each color?
大提示:
每种颜色都至少有三个,所以你可能先抽到每种颜色各三个;下一个就必然形成四个同色
Each color has at least three available, so you could draw three of each before the next one forces a fourth
解答:
最坏情况下,一个人可以先抽到 33 个红球、33 个白球和 33 个蓝球,共 99 个球,还没有四个同色。
第十个球必须与其中某一种颜色相同,从而得到四个同色。因此需要 1010 个球。
所以正确答案是 C。
In the worst case, a person could draw 33 red, 33 white, and 33 blue, which is 99 gumballs, without yet having four of any color.
The next (tenth) gumball must match one of these colors, giving four of that color. So 1010 gumballs are needed.
Thus, the correct answer is C .
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