2021 AMC Fall 10B Problem 3

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

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.

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
$$\frac{1}{2020} + \frac{1}{2021} = \frac{2021 + 2020}{2020 \times 2021} = \frac{4041}{2020 \times 2021}.$$