1951 AMC 12 第 31 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

31.

一次聚会结束时,人们共握手 2828 次。假设每位参加者对其他所有人都同样礼貌,则在场人数为:

A total of 2828 handshakes was exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:

1414

2828

5656

88

77

答案:D
知识点:数对计数二次方程
难度评级:1290
小提示:

若有 nn 人,则每一对人握手一次

With nn people, each unordered pair shakes hands once

大提示:

解方程 (n2)=28\binom n2=28

Solve (n2)=28\binom n2=28

解答:

若有 nn 人,则握手次数为 (n2)=n(n1)2 \binom n2=\frac{n(n-1)}2\text{。}因此 n(n1)=56n(n-1)=56,正解为 n=8n=8

因此,正确答案是 D

With nn people the number of handshakes is (n2)=n(n1)2. \binom n2=\frac{n(n-1)}2. Thus n(n1)=56,n(n-1)=56, and the positive solution is n=8.n=8.

Thus, the correct answer is D.

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