1998 AMC 8 第 25 题

先试着解答 1998 AMC 8 第 25 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 8 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。

25.

三位慷慨的朋友各有一些现金,并按如下方式重新分配:Ami 给 Jan 和 Toy 足够的钱,使他们各自的钱数加倍。然后 Jan 给 Ami 和 Toy 足够的钱,使他们各自的钱数加倍。最后 Toy 给 Ami 和 Jan 足够的钱,使他们各自的钱数加倍。如果 Toy 开始时有 $36\$36,结束时也有 $36\$36,那么三位朋友一共有多少钱?
Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has $36\$36 when they begin and $36\$36 when they end, what is the total amount that all three friends have?
$108\$108
$180\$180
$216\$216
$252\$252
$288\$288
答案:D
知识点:逆推法不变量
小提示:
跟踪 Toy 在前两次分配后的钱数
Track Toy’s amount through the first two exchanges.
大提示:
Toy 最后一次给出的钱数等于 Ami 和 Jan 当时的总钱数
Before Toy’s final turn, the amount he gives equals Ami and Jan’s combined amount.
解答:
Toy 开始有 $36\$36。Ami 使 Toy 的钱加倍后,Toy 有 $72\$72。Jan 再使 Toy 的钱加倍后,Toy 有 $144\$144。
Toy 最后结束时有 $36\$36,所以 Toy 给出了 $144−$36=$108\$144-\$36=\$108。这笔钱使 Ami 和 Jan 的总钱数加倍,因此 Toy 最后行动前 Ami 和 Jan 合计有 $108\$108。
总钱数不变,所以总数是 $144+$108=$252\$144+\$108=\$252。
所以正确答案是 D。
Toy begins with $36.\$36. After Ami doubles Toy’s amount, Toy has $72.\$72. After Jan doubles Toy’s amount, Toy has $144.\$144.
On Toy’s final turn, Toy ends with $36,\$36, so Toy gives away $144−$36=$108.\$144-\$36=\$108. That gift doubles the combined amount of Ami and Jan, so Ami and Jan together had $108\$108 just before Toy’s final turn.
The total amount of money is constant, so the total is $144+$108=$252.\$144+\$108=\$252.
Thus, the correct answer is D .
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.