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

Revision as of 18:27, 15 August 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 answer choice $\boxed{\textbf{(E)}}.$

~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 1

https://youtu.be/ludy6AnQkrI

~Education, the Study of Education

Video Solution by Interstigation

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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