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

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<math>(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041</math>
 
<math>(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041</math>
  
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==Solution==
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==Solution 1==
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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>
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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]]
 
~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]]
  
 +
== Solution 2 ==
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Denote <math>a = 2020</math>. Hence,
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<cmath>
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\begin{align*}
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\frac{2021}{2020} - \frac{2020}{2021}
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& = \frac{a + 1}{a} - \frac{a}{a + 1} \\
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& = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\
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& = \frac{2 a + 1}{a \left( a + 1 \right)} .
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\end{align*}
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</cmath>
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We observe that <math>{\rm gcd} \left( 2a + 1 , a \right) = 1</math> and <math>{\rm gcd} \left( 2a + 1 , a + 1 \right) = 1</math>.
 +
 +
Hence, <math>{\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1</math>.
 +
 +
Therefore, <math>p = 2 a + 1 = 4041</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E) }4041}</math>.
 +
 +
~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
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

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

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