2015 AMC 8 第 13 题

先试着解答 2015 AMC 8 第 13 题,然后核对你的答案与视频与精心整理的解答,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 2015 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。

13.

从集合 {1,2,3,4,5,6,7,8,9,10,11}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} 中可以移除多少个不同的二元素子集,使剩余九个数的平均数为 66?
How many subsets of two elements can be removed from the set {1,2,3,4,5,6,7,8,9,10,11}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} so that the mean (average) of the nine remaining numbers is 6?6?
11
22
33
55
66
答案:D
知识点:平均数系统列举
小提示:
求剩余九个数必须有的总和
Find the sum the remaining nine numbers must have.
大提示:
因此被移除的两个数必须有固定的和
So the two removed numbers must have a fixed sum.
视频讲解:
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文字解答:
原集合的和为 1+2+⋯+11=661+2+\cdots+11=66。移除两个数后剩下 99 个数,平均数要为 66,所以剩余数总和必须是 9⋅6=549\cdot6=54。
因此被移除的两个数之和必须是 66−54=1266-54=12。可能的二元素子集为 {1,11}\{1,11\}、{2,10}\{2,10\}、{3,9}\{3,9\}、{4,8}\{4,8\}、{5,7}\{5,7\},共有 55 种。
所以正确答案是 D。
The original set has sum 1+2+⋯+11=66.1+2+\cdots+11=66. After removing two numbers, 99 numbers remain and must have mean 6,6, so their sum must be 9⋅6=54.9\cdot6=54.
Therefore the two removed numbers must have sum 66−54=12.66-54=12. The possible two-element subsets are {1,11},\{1,11\}, {2,10},\{2,10\}, {3,9},\{3,9\}, {4,8},\{4,8\}, {5,7},\{5,7\}, so there are 55 choices.
Thus, D is the correct answer.
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