Problem 8 · 2020 AMC 8
Medium
Algebra & Patterns
substitutionsum-constraint
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?
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Answer: C — 8072 cents.
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Hint 1 of 2
Don't compute the max and min money separately. Picture turning one penny into a nickel: the coin count stays 2020 but the value jumps by 4 cents. So the difference is just 4 cents per swap, times how many swaps are possible.
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Hint 2 of 2
Total = p + 5n = (p+n) + 4n = 2020 + 4n. The fixed 2020 cancels in the difference, leaving 4 × (range of n).
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Approach: the difference is just (extra value per swap) × (number of swaps)
- Every coin is worth at least 1 cent, so write Total = p + 5n = (p + n) + 4n = 2020 + 4n. The 2020 is fixed; only the 4n changes.
- Constraints “at least one of each” with p + n = 2020 give n from 1 up to 2019.
- Max total has n = 2019, min has n = 1. The fixed 2020 cancels, so the difference is 4(2019 − 1) = 4 × 2018 = 8072.
- You'll see this again as: for “max minus min” questions, peel off the constant part — the spread depends only on what varies. Here each penny→nickel swap adds a constant 4 cents, so the answer is (per-swap gain) × (allowed swaps).
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