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Talk:2020 AMC 8 Problems/Problem 8

Revision as of 23:29, 25 November 2020 by Harvychang (talk | contribs) (→‎Solution 3: new section)
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Solution 3

The smallest amount of money (in cents) that Ricardo can have is $2019$ pennies and $1$ nickel, which is equal to $2019 \cdot 1 + 1 \cdot 5 = 2024$ cents. Similarly, the largest amount of money he can have is $1$ penny and $2019$ nickels, which is equal to $1 \cdot 1 + 2019 \cdot 5 = 10096$ cents. So, the answer is $10096 - 2024 = \boxed{\textbf{(C) }8072}$

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