1998 AMC 12 第 20 题

先试着解答 1998 AMC 12 第 20 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 1998 AMC 12 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。

20.

三张牌背面朝上放在桌上,每张牌上写着一个正整数。凯西、斯泰西和特蕾西得知:
(a)三个数互不相同;
(b)三个数的和为 1313;
(c)三个数从左到右递增。
首先,凯西看了最左边牌上的数,说:“我没有足够的信息确定另外两个数。”接着,特蕾西看了最右边牌上的数,说:“我没有足够的信息确定另外两个数。”最后,斯泰西看了中间牌上的数,说:“我没有足够的信息确定另外两个数。”假设每个人都知道另外两人能够进行完全正确的推理,而且都听到了他们的发言。中间牌上的数是多少?
Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
(a) the numbers are all different,
(b) they sum to 13,13, and
(c) they are in increasing order, left to right.
First, Casey looks at the number on the leftmost card and says, “I don’t have enough information to determine the other two numbers.” Then Tracy looks at the number on the rightmost card and says, “I don’t have enough information to determine the other two numbers.” Finally, Stacy looks at the number on the middle card and says, “I don’t have enough information to determine the other two numbers.” Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?
22
33
44
55
没有足够的信息确定这个数。
There is not enough information to determine the number.
答案:C
知识点:logicordered triples
小提示:
列出所有和为 1313 的递增正整数三元组
List all increasing positive triples with sum 1313
大提示:
按照三句话的先后顺序排除三元组,并利用每位发言者已经得知的信息
Eliminate triples in the order of the three statements, using what each speaker has learned
解答:
八个三元组为 (1,2,10),(1,3,9),(1,4,8),(1,5,7),(2,3,8),(2,4,7),(2,5,6),(3,4,6)。 \begin{gathered} (1,2,10),(1,3,9),\\ (1,4,8),(1,5,7),\\ (2,3,8),(2,4,7),\\ (2,5,6),(3,4,6) \end{gathered}\text{。}凯西的话排除 (3,4,6)(3,4,6)。得知这一点后,特蕾西的话排除最右端为 10,910,9 或 66 的情形。剩下的三元组为 (1,4,8),(2,3,8)(1,4,8),(2,3,8) 和 (1,5,7),(2,4,7)(1,5,7),(2,4,7)。若中间数为 33 或 55,斯泰西此时就能确定三元组,所以她的话迫使中间数为 44。因此正确答案是 C。
The eight triples are (1,2,10),(1,3,9),(1,4,8),(1,5,7),(2,3,8),(2,4,7),(2,5,6),(3,4,6). \begin{gathered} (1,2,10),(1,3,9),\\ (1,4,8),(1,5,7),\\ (2,3,8),(2,4,7),\\ (2,5,6),(3,4,6). \end{gathered} Casey’s statement excludes (3,4,6).(3,4,6). With that known, Tracy’s statement excludes rightmost values 10,9,10,9, and 6.6. The remaining triples are (1,4,8),(2,3,8)(1,4,8),(2,3,8) and (1,5,7),(2,4,7).(1,5,7),(2,4,7). A middle value 33 or 55 would now identify the triple, so Stacy’s statement forces the middle value 4.4. Thus C is correct.
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