Difference between revisions of "2020 AMC 8 Problems/Problem 8"
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The greatest amount of money will occur when he has the greatest number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is 2019. This leads to a total of <math>(2019\cdot 5 + 1)</math> cents. The least amount of money will occur when he has the greatest number of pennies. Since he must have at least one nickel, the greatest number of pennies he can have is 2019. This leads to a total of <math>(2019\cdot 1 + 5)</math> cents. Thus, the difference between the greatest possible amount of money and the least possible amount of money is <math>(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4(2019-1)=4\cdot 2018=8072\implies\boxed{\textbf{(C) }8072}</math>.<br> | The greatest amount of money will occur when he has the greatest number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is 2019. This leads to a total of <math>(2019\cdot 5 + 1)</math> cents. The least amount of money will occur when he has the greatest number of pennies. Since he must have at least one nickel, the greatest number of pennies he can have is 2019. This leads to a total of <math>(2019\cdot 1 + 5)</math> cents. Thus, the difference between the greatest possible amount of money and the least possible amount of money is <math>(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4(2019-1)=4\cdot 2018=8072\implies\boxed{\textbf{(C) }8072}</math>.<br> | ||
~ junaidmansuri | ~ junaidmansuri | ||
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| + | ==Solution 2== | ||
| + | |||
| + | The idea is maximizing the number of nickels and then maximizing the number of pennies, and then take their difference. This is given by <math>(2019\cdot 5+1)-(2019\cdot 1+5)=\textbf{(C) }8072</math>. | ||
| + | |||
| + | -franzliszt | ||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=7|num-a=9}} | {{AMC8 box|year=2020|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:38, 18 November 2020
Contents
Problem 8
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
The greatest amount of money will occur when he has the greatest number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is 2019. This leads to a total of 
cents. The least amount of money will occur when he has the greatest number of pennies. Since he must have at least one nickel, the greatest number of pennies he can have is 2019. This leads to a total of 
cents. Thus, the difference between the greatest possible amount of money and the least possible amount of money is 
.
~ junaidmansuri
Solution 2
The idea is maximizing the number of nickels and then maximizing the number of pennies, and then take their difference. This is given by 
.
-franzliszt
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. 
