2020 AMC 8 Problems/Problem 8

Problem 8

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?

$\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$

Solution 1

The greatest amount of money will occur when he has the greatest number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is 2019. This leads to a total of $(2019\cdot 5 + 1)$ cents. The least amount of money will occur when he has the greatest number of pennies. Since he must have at least one nickel, the greatest number of pennies he can have is 2019. This leads to a total of $(2019\cdot 1 + 5)$ cents. Thus, the difference between the greatest possible amount of money and the least possible amount of money is $(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4(2019-1)=4\cdot 2018=8072\implies\boxed{\textbf{(C) }8072}$ .
~ junaidmansuri

Solution 2

The idea is maximizing the number of nickels and then maximizing the number of pennies, and then take their difference. This is given by $(2019\cdot 5+1)-(2019\cdot 1+5)=\textbf{(C) }8072$ .

-franzliszt

2020 AMC 8 (Problems • Answer Key • Resources)
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2020 AMC 8 Problems/Problem 8

Contents

Problem 8

Solution 1

Solution 2

See also