2021 AMC Fall 10B Problem 3

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

$\frac{a}{b} - \frac{b}{a} = \frac{a^2 - b^2}{ab} = \frac{(a-b)(a+b)}{ab}$, and when $a = b + 1$ this is $\frac{a + b}{ab}$. The reduced-form check uses the Euclidean idea: $\gcd(a+b, a) = \gcd(b, a)$, which is $1$ for consecutive integers.

Write $\frac{2021}{2020} = 1 + \frac{1}{2020}$ and $\frac{2020}{2021} = 1 - \frac{1}{2021}$. Their difference is $\frac{1}{2020} + \frac{1}{2021} = \frac{2021 + 2020}{2020 \cdot 2021}$, numerator $4041$.

• Stopping at the unsimplified numerator 2021^2 - 2020^2 without recognizing the difference of squares, or assuming the fraction must reduce further.
• Subtracting in the wrong order and reporting $p$ from $\frac{2020}{2021} - \frac{2021}{2020}$, which is negative.