2020 AMC 8 Problems/Problem 8
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (🚀Very Fast🚀)
- 7 Video Solution by North America Math Contest Go Go Go
- 8 Video Solution by WhyMath
- 9 Video Solution
- 10 Video Solution by Interstigation
- 11 Video Solution by STEMbreezy
- 12 See also
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 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 him 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
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=5OsfZIo0eURrtNri&t=845
~Math-X
Video Solution (🚀Very Fast🚀)
~Education, the Study of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=w4S6K9XbHJg
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=300
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS&t=181
~STEMbreezy
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.