Question

$A$ 先给 $B$ 与 $B$ 原有钱数相同的美分，再给 $C$ 与 $C$ 原有钱数相同的美分。接着， $B$ 同样分别给 $A$ 和 $C$ 与他们当时各自钱数相同的美分。最后， $C$ 也同样分别给 $A$ 和 $B$ 与他们当时各自钱数相同的美分。如果最终每人都有 $16$ 美分，那么 $A$ 最初有多少美分？

$A$ gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?

$24$

$26$

$28$

$30$

$32$

答案：B

知识点：钱币逆推法

难度评级：1780

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。 $C$ 给钱之前，三人的钱数为 $(8,8,32)$ 。 $B$ 给钱之前，钱数为 $(4,28,16)$ 。 $A$ 给钱之前， $B$ 和 $C$ 必须分别有 $14$ 和 $8$ 美分，而 $A$ 有 $4+14+8=26$ 美分。

所以正确答案是 B。

Reverse the transactions. Before $C$ gives, the holdings are $(8,8,32).$ Before $B$ gives, they are $(4,28,16).$ Before $A$ gives, $B$ and $C$ must have had $14$ and $8,$ while $A$ had $4+14+8=26.$

Therefore, the correct answer is B.

1963 AMC 12 第 23 题

答案：B

难度评级：1780

小提示：

大提示：

解答：

其他年份的第 23 题