# 2021 Fall AMC 10B Problems/Problem 3

## 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 $\boxed{\textbf{(E) }4041}$.

## 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

~Education, the Study of Everything

~savannahsolver

## Video Solution by TheBeautyofMath

~IceMatrix

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