Question

Ricardo 有 $2020$ 枚硬币，其中一些是 $1$ 分硬币，其余是 $5$ 分硬币。他至少有一枚一分硬币，也至少有一枚五分硬币。Ricardo 可能拥有的钱数最大值与最小值相差多少分？

Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -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?

$8062$

$8068$

$8072$

$8076$

$8082$

答案：C

知识点：一次方程最优化

难度评级：1020

视频讲解：

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文字解答：

最大金额对应 $2019$ 枚五美分硬币和 $1$ 枚一美分硬币；最小金额对应 $1$ 枚五美分硬币和 $2019$ 枚一美分硬币。

从第二种情况变到第一种情况，共把 $2018$ 枚一美分硬币换成五美分硬币。每次替换增加 $5-1=4$ 美分，所以差为 $2018\cdot4=8072\text{ cents}.$

所以正确答案是 C。

The greatest value occurs with $2019$ nickels and $1$ penny, while the least occurs with $1$ nickel and $2019$ pennies.

Between these two cases, $2018$ pennies have been replaced by nickels. Each replacement adds $5-1=4$ cents, so the difference is $2018\cdot4=8072\text{ cents}.$

Thus, the correct answer is C.

2020 AMC 8 第 8 题

答案：C

难度评级：1020

视频讲解：

文字解答：

其他年份的第 8 题