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Difference between revisions of "2020 AMC 8 Problems/Problem 8"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
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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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==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!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==

Latest revision as of 06:17, 24 January 2024

Problem

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 amounts of money that Ricardo can have?

$\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$

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 $2019$, giving a total of $(2019\cdot 5 + 1)$ 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 $2019$, giving him a total of $(2019\cdot 1 + 5)$ cents. Hence the required difference is \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{\textbf{(C) }8072}\]

Solution 2

Suppose Ricardo has $p$ pennies, so then he has $(2020-p)$ nickels. In order to have at least one penny and at least one nickel, we require $p \geq 1$ and $2020 - p \geq 1$, i.e. $1 \leq p \leq 2019$. The number of cents he has is $p+5(2020-p) = 10100-4p$, so the maximum is $10100-4 \cdot 1$ and the minimum is $10100 - 4 \cdot 2019$, and the difference is therefore \[(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{\textbf{(C) }8072}\]

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🚀)

https://youtu.be/SkEGB5NSjxU

~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

https://youtu.be/OWNZBPIzQXQ

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8

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

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