1963 AMC 12 第 23 题

先试着解答 1963 AMC 12 第 23 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 1963 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。

23.

AA 先给 BB 与 BB 原有钱数相同的美分,再给 CC 与 CC 原有钱数相同的美分。接着,BB 同样分别给 AA 和 CC 与他们当时各自钱数相同的美分。最后,CC 也同样分别给 AA 和 BB 与他们当时各自钱数相同的美分。如果最终每人都有 1616 美分,那么 AA 最初有多少美分?
AA gives BB as many cents as BB has and CC as many cents as CC has. Similarly, BB then gives AA and CC as many cents as each then has. CC, similarly, then gives AA and BB as many cents as each then has. If each finally has 1616 cents, with how many cents does AA start?
2424
2626
2828
3030
3232
答案:B
知识点:钱币逆推法
小提示:
从最终每人的钱数倒推
Work backward from the final holdings
大提示:
某人给钱之前,每位收款人的钱数应为收款后的一半
Immediately before someone gives, each recipient must have half of the amount held just after that gift
解答:
逆向还原这些交易。CC 给钱之前,三人的钱数为 (8,8,32)(8,8,32)。BB 给钱之前,钱数为 (4,28,16)(4,28,16)。AA 给钱之前,BB 和 CC 必须分别有 1414 和 88 美分,而 AA 有 4+14+8=264+14+8=26 美分。
所以正确答案是 B。
Reverse the transactions. Before CC gives, the holdings are (8,8,32).(8,8,32). Before BB gives, they are (4,28,16).(4,28,16). Before AA gives, BB and CC must have had 1414 and 8,8, while AA had 4+14+8=26.4+14+8=26.
Therefore, the correct answer is B.
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