AMC8 2020

AMC8 2020 · Q8

AMC8 2020 · Q8. It mainly tests Money / coins.

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 分硬币），其余是镍币（5 分硬币）。他至少有一枚便士和一枚镍币。Ricardo 可能拥有的最大金额和最小金额之间的差值是多少分？

(A) \text{806} \text{806}

(B) \text{8068} \text{8068}

(D) \text{8076} \text{8076}

(E) \text{8082} \text{8082}

Answer

Correct choice: (C)

正确答案：（C）

Solution

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 $(2019\cdot 5 + 1)$ 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 $(2019\cdot 1 + 5)$ cents. Hence the required difference is \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{\textbf{(C) }8072}\]

显然，当他拥有最大数量镍币时，Ricardo 的金额最大。由于他必须至少有一枚便士，他最多可有 $2019$ 枚镍币，总额为 $(2019\cdot 5 + 1)$ 分。类似地，当他拥有最大数量便士时，金额最小；由于他必须至少有一枚镍币，他最多可有 $2019$ 枚便士，总额为 $(2019\cdot 1 + 5)$ 分。因此所需差值为 \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{\textbf{(C) }8072}\]

Topics

AR.014 Money / coins