For any real number $x$, we let $\lfloor x \rfloor$ be the unique integer $n$ such that $n \leq x < n+1$. For example. $\lfloor 31.415 \rfloor = 31$. Compute \[2020^{2021} - \left\lfloor\frac{2020^{2021}}{2021} \right  \rfloor (2021).\]


[i]2021 CCA Math Bonanza Team Round #3[/i]
1. Let \( x = \frac{2020^{2021}}{2021} \). We need to compute \( 2020^{2021} - \left\lfloor x \right\rfloor \cdot 2021 \).

2. First, observe that:
   \[
   2020^{2021} = 2021 \cdot x
   \]
   This follows from the definition of \( x \).

3. Next, we need to find \( \left\lfloor x \right\rfloor \). Since \( x = \frac{2020^{2021}}{2021} \), we can approximate \( x \) by considering the binomial expansion:
   \[
   2020^{2021} = (2021 - 1)^{2021}
   \]
   Using the binomial theorem, we have:
   \[
   (2021 - 1)^{2021} = \sum_{k=0}^{2021} \binom{2021}{k} \cdot 2021^{2021-k} \cdot (-1)^k
   \]
   The dominant term here is \( 2021^{2021} \), and the next significant term is \( -2021^{2020} \). For large \( n \), the higher-order terms become negligible.

4. Therefore, we approximate:
   \[
   2020^{2021} \approx 2021^{2021} - 2021^{2020}
   \]
   Dividing by 2021, we get:
   \[
   x \approx 2021^{2020} - 2021^{2019}
   \]
   Since \( x \) is very close to an integer, we can write:
   \[
   x = a - \frac{1}{2021}
   \]
   for some integer \( a \).

5. Thus, we have:
   \[
   \left\lfloor x \right\rfloor = x - \frac{2020}{2021}
   \]

6. Now, we compute:
   \[
   2020^{2021} - \left( x - \frac{2020}{2021} \right) \cdot 2021
   \]
   Simplifying, we get:
   \[
   2020^{2021} - \left( \frac{2020^{2021}}{2021} - \frac{2020}{2021} \right) \cdot 2021
   \]
   \[
   = 2020^{2021} - \left( 2020^{2021} - 2020 \right)
   \]
   \[
   = 2020
   \]

The final answer is \(\boxed{2020}\)