AMC 8 1998 Problem 25
AMC8 1998 年第 25 题
方程与不等式★★★★☆
Three generous friends, each with some money, redistribute the money as followed: Amy gives enough money to Jan and Toy to double each of their amounts. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
三位慷慨的朋友各自拥有一些钱，按如下方式重新分配：Amy 给 Jan 和 Toy 足够的钱，使他们各自的金额翻倍。然后 Jan 给 Amy 和 Toy 足够的钱，使他们的金额翻倍。最后，Toy 给 Amy 和 Jan 足够的钱，使他们的金额翻倍。如果 Toy 一开始有 36 美元，最后也有 36 美元，那么三位朋友共有多少钱？
- A.108
- B.180
- C.216
- D.252
- E.288
正确答案 · Correct Answer
D
解析 · Solution
解法1
常规解法 · 盯住 Toy 顺推
每次转账总额不变。Toy 起始 36，Amy 行动后翻倍到 72，Jan 行动后再翻倍到 144。轮到 Toy 时它把别人各翻倍、自己最后剩 36，说明它送出 144-36=108，正好等于此前 Amy 与 Jan 之和。故总额 =108+144=252。
巧解 · 总额守恒 + 末态
设总额恒为 。最后 Amy、Jan 刚被翻倍、Toy 为 36；Toy 在最后一步前持有 144（两次翻倍），送出 144-36=108 即翻倍前的 Amy+Jan，翻倍后变 216，故 T=216+36=252。
Solution 1
Elementary · follow Toy forward
The total amount is unchanged by the transfers. Toy starts at 36, is doubled to 72 after Amy's move, and doubled again to 144 after Jan's move. On Toy's move Toy doubles the others and ends at 36, so Toy gives away 144-36=108, which equals Amy+Jan just before that move. Hence the total is 108+144=252.
Slick · total conserved + final state
Let the total be (constant). At the end Amy and Jan have just been doubled and Toy has 36. Before Toy's move Toy held 144 (two doublings of 36), so Amy+Jan was 144-36=108; doubling makes it 216, and T=216+36=252.
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