AMC 8 2020 Problem 8
AMC8 2020 年第 8 题
表达式与运算★★☆☆☆
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 amounts of money that Ricardo can have?
Ricardo 有 2020 枚硬币，其中一些是便士（1 美分硬币），其余的是镍币（5 美分硬币）。他有至少一枚一分便士和一枚五分镍币。Ricardo 可能拥有的最大金额和可能拥有的最小金额相差多少美分？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is 2019, giving a total of cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies; as he must have at least one nickel, the greatest number of pennies he can have is also 2019, giving him a total of cents. Hence the required difference is
Suppose Ricardo has pennies, so then he has (2020-p) nickels. In order to have at least one penny and at least one nickel, we require and , i.e. . The number of cents he has is p+5(2020-p) = 10100-4p, so the maximum is and the minimum is , and the difference is therefore
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