2020 AMC 8 Problems/Problem 8
Problem
Ricardo has coins, some of which are pennies (-cent coins) and the rest of which are nickels (-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?
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 , giving a total of 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 , giving a total of cents. Hence the required difference is
Solution 2
Suppose Ricardo has pennies, so then he has nickels. In order to have at least one penny and at least one nickel, we require and , i.e. . The number of cents he has is , so the maximum is and the minimum is , and the difference is therefore
Solution 3
The smallest amount of money (in cents) that Ricardo can have is pennies and nickel, which is cents. Similarly, the largest amount of money he can have is penny and nickels, which is cents. So, the answer is
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.