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2020 AMC 8 Problems/Problem 8

Revision as of 06:43, 20 November 2020 by Sevenoptimus (talk | contribs) (Added an additional solution, removed a duplicate solution, and fixed grammar/wording)

Problem

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

Clearly, the amount of money he 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 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 he has $p$ pennies, so 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 $(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}$.

Video Solution

https://youtu.be/OWNZBPIzQXQ

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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