2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 8 Easy

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?

  • A.

    806

  • B.

    8068

  • C.

    8072

  • D.

    8076

  • E.

    8082

Answer:C

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 (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}

 

Solution 2

Suppose Ricardo has p pennies, so then he has (2020-p) nickels. In order to have at least one penny and at least one nickel, we require p \geq 1 and 2020 - p \geq 1, i.e. 1 \leq p \leq 2019. The number of cents he has is p+5(2020-p) = 10100-4p, so the maximum is 10100-4 \cdot 1 and the minimum is 10100 - 4 \cdot 2019, and the difference is therefore

(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}

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