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第11届美国数学竞赛于1982年5月4日举行。下面是这届竞赛的试题及我们所拟的一份解答。限于水平,不妥之处,祈请读者指正。 1.一次集会有1982个人参加,其中任意的四个人中至少有一个人认识其余的三个人。问在这次集会上,认识全体到会者的人至少有多少位? 解我们证明认识全体到会者的人至少有1979名。换句话说,至多有三个人不认识全体到会者。采用反证法。假设至少有四个人不认识全体到会者。A为其中之一,A不认识B。这时,除A、B两人外,还有C不认识全体到会者。假如C不认识D,并且D不是A、B,那么在A、B、C、D这四个人中,每一个都不全认识其余的三个人,与已知矛盾。因此C不认识的人一定是A或B。这时,除A、B、C、三个人外,还有D不认识全体到会者。根据刚才对C的同样的推理,D所不认识的人一定是A、
The 11th American Mathematical Contest was held on May 4, 1982. Here are the questions for this competition and the answers we have proposed. Limited to the level, inappropriate, pray readers correct me. 1. There are 1982 individuals participating in a rally. At least one of the four persons involved knows the remaining three. Asked at this rally, at least how many people knew all the attendees? We have at least 1979 people who understand that we have known all the attendees. In other words, at most three people do not know the whole audience. Use counter-evidence. Assume that at least four people do not recognize the attendees. A is one of them, A does not know B. At this time, in addition to A and B, there were C who did not recognize the attendees. If C does not recognize D, and D is not A, B, then each of the four persons A, B, C, and D does not fully recognize the remaining three persons, contradicting the known ones. So people who C does not know must be A or B. At this time, in addition to A, B, C, and three people, there are D who do not recognize the attendees. According to the same reasoning that was just made for C, people who D did not know must be A,