2008 AMC 8 第 24 题

先试着解答 2008 AMC 8 第 24 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 2008 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。

24.

编号为 11 到 1010 的十块牌面朝下放着。随机翻开一块牌,并掷一个骰子。牌上的数与骰子点数的乘积是平方数的概率是多少?
Ten tiles numbered 11 through 1010 are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
 110\ \dfrac{1}{10}
 16\ \dfrac{1}{6}
 1160\ \dfrac{11}{60}
 15\ \dfrac{1}{5}
 730\ \dfrac{7}{30}
答案:C
知识点:基本概率完全平方数分类讨论
小提示:
共有 10⋅610\cdot 6 个等可能的牌和骰子组合
There are 10⋅610\cdot 6 equally likely tile-die pairs.
大提示:
对每个骰子点数,数一数哪些牌号能使乘积成为平方数
For each die roll, count tile numbers that make the product a square.
解答:
如果骰子点数为 11,牌可以是 11、44、99,有 33 种组合。
如果骰子点数为 22,牌可以是 22、88,有 22 种组合。
如果骰子点数为 33,牌可以是 33,有 11 种组合。
如果骰子点数为 44,牌可以是 11、44、99,有 33 种组合。
如果骰子点数为 55,牌可以是 55,有 11 种组合。
如果骰子点数为 66,牌可以是 66,有 11 种组合。
组合总数为 3+2+1+3+1+1=11。3+2+1+3+1+1=11\text{。}共有 6060 个等可能组合,所以概率是 1160\dfrac{11}{60} 。
所以正确答案是 C。
If the rolled number was 1,1, then the tile can be 1,1, 4,4, 9,9, yielding 33 combinations.
If the rolled number was 2,2, then the tile can be 2,2, 8,8, yielding 22 combinations.
If the rolled number was 3,3, then the tile can be 3,3, yielding 11 combination.
If the rolled number was 4,4, then the tile can be 1,1, 4,4, 9,9, yielding 33 combinations.
If the rolled number was 5,5, then the tile can be 5,5, yielding 11 combination.
If the rolled number was 6,6, then the tile can be 6,6, yielding 11 combination.
The total number of combinations is 3+2+1+3+1+1=11.3+2+1+3+1+1=11. There are 6060 combinations each with equal likelihood, so the probability is 1160.\dfrac{11}{60} .
Thus, the answer is C .
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