Difference between revisions of "2021 Fall AMC 10B Problems/Problem 3"

Latest revision as of 23:53, 29 December 2022

Problem

The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$

$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$

Solution 1

We write the given expression as a single fraction: $\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}$ by cross multiplication. Then by factoring the numerator, we get $\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.$ The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\boxed{\textbf{(E) }4041}$ .

~Aops-g5-gethsemanea2

Solution 2

Denote $a = 2020$ . Hence, $\begin{align*} \frac{2021}{2020} - \frac{2020}{2021} & = \frac{a + 1}{a} - \frac{a}{a + 1} \\ & = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\ & = \frac{2 a + 1}{a \left( a + 1 \right)} . \end{align*}$

We observe that ${\rm gcd} \left( 2a + 1 , a \right) = 1$ and ${\rm gcd} \left( 2a + 1 , a + 1 \right) = 1$ .

Hence, ${\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1$ .

Therefore, $p = 2 a + 1 = 4041$ .

Therefore, the answer is $\boxed{\textbf{(E) }4041}$ .

~Steven Chen (www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=160

Video Solution

https://youtu.be/ludy6AnQkrI

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/PPIZH_iBTJw

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/lC7naDZ1Eu4?t=378

~IceMatrix

@@ Line 6: / Line 6: @@
 ==Solution 1==
-We write the given expression as a single fraction: <cmath>\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}</cmath> by cross multiplication. Then by factoring the numerator, we get <cmath>\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.</cmath> The question is asking for the numerator, so our answer is <math>2021+2020=4041,</math> giving answer choice <math>\boxed{\textbf{(E)}}.</math>
+We write the given expression as a single fraction: <cmath>\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}</cmath> by cross multiplication. Then by factoring the numerator, we get <cmath>\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.</cmath> The question is asking for the numerator, so our answer is <math>2021+2020=4041,</math> giving <math>\boxed{\textbf{(E) }4041}</math>.
 ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
@@ Line 34: / Line 34: @@
 https://youtu.be/p9_RH4s-kBA?t=160
+==Video Solution==
+https://youtu.be/ludy6AnQkrI
+~Education, the Study of Everything
+==Video Solution by WhyMath==
+https://youtu.be/PPIZH_iBTJw
+~savannahsolver
+==Video Solution by TheBeautyofMath==
+https://youtu.be/lC7naDZ1Eu4?t=378
+~IceMatrix
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
 {{MAA Notice}}

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 AMC 10 Problems and Solutions

Page

Toolbox

Search