2020 AMC 8 第 8 题

先试着解答 2020 AMC 8 第 8 题,然后核对你的答案与视频与精心整理的解答,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 2020 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。

8.

里卡多有 20202020 枚硬币,其中一些是 11 美分硬币,其余是 55 美分硬币。他至少有一枚一美分硬币,也至少有一枚五美分硬币。里卡多可能拥有的钱数最大值与最小值相差多少美分?
Ricardo has 20202020 coins, some of which are pennies (11-cent coins) and the rest of which are nickels (55-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?
80628062
80688068
80728072
80768076
80828082
答案:C
知识点:一次方程最优化
小提示:
先保证每种硬币至少留一枚
Leave one coin of each type fixed
大提示:
把一枚一分硬币换成一枚五分硬币会增加 44 分
Changing one penny to one nickel changes the value by 44 cents
视频讲解:
解答视频缩略图
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文字解答:
最大金额对应 20192019 枚五美分硬币和 11 枚一美分硬币;最小金额对应 11 枚五美分硬币和 20192019 枚一美分硬币。
从第二种情况变到第一种情况,共把 20182018 枚一美分硬币换成五美分硬币。每次替换增加 5−1=45-1=4 美分,所以差为 2018⋅4=8072 2018\cdot4=8072 美分。
所以正确答案是 C。
The greatest value occurs with 20192019 nickels and 11 penny, while the least occurs with 11 nickel and 20192019 pennies.
Between these two cases, 20182018 pennies have been replaced by nickels. Each replacement adds 5−1=45-1=4 cents, so the difference is 2018⋅4=8072 2018\cdot4=8072 cents.
Thus, the correct answer is C.
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