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?

A. 8062

B. 8068

C. 8072

D. 8076

E. 8082

1. 2020 AMC 8 官方解析

To obtain the greatest possible amount of money, use as many nickels as possible: $2019$ nickels and $1$ penny. This gives $2019 \cdot 5 + 1 \cdot 1 = 10,096$ cents. Similarly, to obtain the least possible amount of money, use as many pennies as possible: $1$ nickel and $2019$ pennies.
This gives $1 \cdot 5 + 2019 \cdot 1 = 2024$ cents. Therefore the difference between the greatest possible and least possible amounts of money is $10,096 - 2024 = 8072$ cents.
为了获得最大的金额，要使$5$美分硬币尽可能的多：$2019$枚$5$美分硬币和$1$枚$1$美分硬币. 金额中共为 $2019 \cdot 5 + 1 \cdot 1 = 10,096$ 美分. 同样的，要获得最小的金额，应该有尽可能多的$1$美分硬币：$1$枚$1$美元的硬币和 $2019$ 枚$5$美元硬币. 总金额为$1 \cdot 5 + 2019 \cdot 1 = 2024$美分. 因此，最大可能金额与最小可能金额之差为 $10,096 - 2024 = 8072$ 美分.
One penny and one nickel will be used in either case so they can be removed, leaving $2018$ coins. The greatest possible sum will result from having $2018$ nickels, and the least possible sum will result from having $2018$ pennies. The difference in amounts is $2018 \cdot 5 - 2018 \cdot 1 = 2018 \cdot 4 = 8072$ cents.
获得最大金额和最小金额两种情况下都会有一枚$1$美分硬币和一枚$5$美分硬币，故$2020$枚硬币中必须有一枚$1$美分硬币和一枚$5$美分硬币. 剩下 $2018$ 枚硬币. 具有 $2018$ 枚$5$美分硬币将使其获得最大的可能金额，而具有 $2018$ 枚$1$美分硬币将使其获得最小的可能金额. 故最大可能金额与最小可能金额之差为 $2018 \cdot 5 - 2018 \cdot 1 = 2018 \cdot 4 = 8072$ 美分.