Difference between revisions of "2020 AMC 8 Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
Suppose Ricardo has <math>p</math> pennies, so then he has <math>(2020-p)</math> nickels. In order to have at least one penny and at least one nickel, we require <math>p \geq 1</math> and <math>2020 - p \geq 1</math>, i.e. <math>1 \leq p \leq 2019</math>. The number of cents he has is <math>p+5(2020-p) = 10100-4p</math>, so the maximum is <math>10100-4 \cdot 1</math> and the minimum is <math>10100 - 4 \cdot 2019</math>, and the difference is therefore <cmath>(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}</cmath> | Suppose Ricardo has <math>p</math> pennies, so then he has <math>(2020-p)</math> nickels. In order to have at least one penny and at least one nickel, we require <math>p \geq 1</math> and <math>2020 - p \geq 1</math>, i.e. <math>1 \leq p \leq 2019</math>. The number of cents he has is <math>p+5(2020-p) = 10100-4p</math>, so the maximum is <math>10100-4 \cdot 1</math> and the minimum is <math>10100 - 4 \cdot 2019</math>, and the difference is therefore <cmath>(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}</cmath> | ||
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| + | == Solution 3 == | ||
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| + | The smallest amount of money (in cents) that Ricardo can have is <math>2019</math> pennies and <math>1</math> nickel, which is equal to <math>2019 \cdot 1 + 1 \cdot 5 = 2024</math> cents. Similarly, the largest amount of money he can have is <math>1</math> penny and <math>2019</math> nickels, which is equal to <math>1 \cdot 1 + 2019 \cdot 5 = 10096</math> cents. So, the answer is <math>10096 - 2024 = \boxed{\textbf{(C) }8072}</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:29, 25 November 2020
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 ![\[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{\textbf{(C) }8072}\]](http://latex.artofproblemsolving.com/7/a/a/7aad883b32b506008cba937df84f736d314828c0.png)
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 ![\[(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}\]](http://latex.artofproblemsolving.com/6/8/6/68604d817623e85842cbf1a05bab0254725f67d3.png)
Solution 3
The smallest amount of money (in cents) that Ricardo can have is pennies and nickel, which is equal to 
cents. Similarly, the largest amount of money he can have is penny and nickels, which is equal to 
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. 
