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2021 AMC Fall 10B Problem 3

2021 AMC Fall 10B Problem 3AMC 10 Step by Step

Name: Date: October 28, 2026

Algebraic Manipulation · difficulty 1 of 5 · about 2 min · position P1-10

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 $p?$

A) $1$
B) $9$
C) $2020$
D) $2021$
E) $4041$

Answer

Key insight

Over a common denominator the top becomes 2021 squared minus 2020 squared, which factors as (1)(4041). Nothing cancels, so p is 4041.

Solution

1. Common denominator. To subtract fractions, give them the same bottom, $2020 \times 2021$ :
$\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021 \times 2021 - 2020 \times 2020}{2020 \times 2021}.$
2. Simplify the top without big multiplication. The top is a difference of two squares, and $a^2 - b^2 = (a - b)(a + b)$ :
$2021^2 - 2020^2 = (2021 - 2020)(2021 + 2020) = 1 \times 4041 = 4041.$
3. Check for cancelling. The fraction is now $\dfrac{4041}{2020 \times 2021}$ . Could $4041$ share a factor with $2020$ ?
- Notice $4041 = 2020 + 2021$ .
- If some number divided both $4041$ and $2020$ , it would also divide their difference, $4041 - 2020 = 2021$ .
- But $2020$ and $2021$ are next to each other, and two numbers next to each other share no factor bigger than $1$ .
- The same argument shows $4041$ and $2021$ share no factor.
4. So nothing cancels, and $p = 4041$ .

The answer is $\boxed{\textbf{(E)}\ 4041}$ .

Why this works

For any two numbers, $\frac{a}{b} - \frac{b}{a} = \frac{a^2 - b^2}{ab} = \frac{(a - b)(a + b)}{ab}$ . When $a$ and $b$ are next to each other, $a - b = 1$ , and the top is just $a + b$ . Whenever you see a square minus a square, factor it instead of multiplying out.

Alternative approach

Split each fraction into $1$ plus a little bit:
$\frac{2021}{2020} = 1 + \frac{1}{2020}, \qquad \frac{2020}{2021} = 1 - \frac{1}{2021}.$
Subtracting, the $1$ s cancel and you get
$\frac{1}{2020} + \frac{1}{2021} = \frac{2021 + 2020}{2020 \times 2021} = \frac{4041}{2020 \times 2021}.$

The trap

Grinding out 2021 squared and 2020 squared by hand and slipping, or assuming the fraction must simplify further.

Concise solution

Put the two fractions over the common denominator $2020 \cdot 2021$ :
$\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021^2 - 2020^2}{2020 \cdot 2021}.$
The numerator is a difference of squares:
$2021^2 - 2020^2 = (2021 - 2020)(2021 + 2020) = 1 \cdot 4041 = 4041.$

Is the fraction $\frac{4041}{2020 \cdot 2021}$ already in lowest terms? Since $4041 = 2020 + 2021$ , any common factor of $4041$ and $2020$ would also divide $4041 - 2020 = 2021$ , but consecutive integers $2020$ and $2021$ are coprime. The same argument works for $4041$ and $2021$ . So $p = 4041$ .

The answer is $\boxed{\textbf{(E)}\ 4041}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Number Theory, Chapter 4: Prime Factorization (§4.7)

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta