20. Ana plays with 𝑛 × 𝑛 boards by placing a token in each of the cells with no common points with other cells containing tokens. In the picture beside we see how to place as many chips as possible on 5 × 5 and 6 × 6 boards. In this way, how many chips can Ana possibly put on a 2020 × 2020 board? (A) 2020 (B) 4039 (C) 674 ଶ (D) 1010 ଶ (E) 2020 ଶ

To solve this problem, we need to determine how Ana can place the maximum number of tokens on a 2020 × 2020 board such that no two tokens share a row, column, or diagonal. This is a classic problem of placing tokens on a board such that they are not attacking each other, similar to the n-queens problem but simplified to exclude diagonals.

Ana must place the tokens in such a way that they form a checkerboard pattern to maximize the number of tokens. Each token should be placed in alternating rows and columns. This means that if a token is placed in a particular cell, the adjacent cells (horizontally, vertically, and diagonally) cannot contain any tokens.

To determine how many tokens can be placed, consider a checkerboard pattern where tokens occupy either all the black squares or all the white squares. For an \( n \times n \) board, half of the squares (approximately) can be filled in this pattern.

**Solution:**

1. **Checkerboard Pattern Approach:**
   - A checkerboard pattern on an \( n \times n \) board divides the board into two equal parts of approximately \( \frac{n^2}{2} \) squares each.
   - For a 2020 × 2020 board, the total number of squares is \( 2020^2 \).

2. **Calculating the Maximum Number of Tokens:**
   - The maximum number of tokens Ana can place is approximately half the total number of squares on the board because of the checkerboard pattern.
   - Hence, the maximum number of tokens is:
     \[
     \text{Maximum Tokens} = \left\lfloor \frac{2020^2}{2} \right\rfloor
     \]
   - Calculating this:
     \[
     2020^2 = 4080400
     \]
     \[
     \frac{4080400}{2} = 2040200
     \]

Therefore, Ana can place a maximum of **2,040,200** tokens on a 2020 × 2020 board, which matches option (E) \(2020^2\).

The correct answer is therefore **(E) 2020^2**.

Would you like more details or have any questions about this problem?

Here are some related questions to expand your understanding:
1. How does the placement strategy change if the board size is an odd number, like 2021 × 2021?
2. What if the tokens can attack diagonally, how would the strategy change?
3. What is the maximum number of tokens on an 8 × 8 board using this method?
4. How does this problem relate to the n-queens problem?
5. How many tokens can be placed on a 7 × 7 board using the same strategy?
6. What is the general formula for the maximum number of tokens on an \( n \times n \) board?
7. How does the number of tokens change if the board is rectangular, such as 2020 × 2021?
8. Can this method be adapted to three-dimensional boards?

**Tip:** In problems involving grids, consider symmetry and patterns such as checkerboards to simplify and optimize arrangements.

Learn how to maximize the placement of tokens on a 2020 × 2020 board without any tokens sharing a row, column, or diagonal. Explore the combinatorial strategy resembling a checkerboard pattern and calculate the maximum number of tokens possible. Suitable for high school students interested in combinatorial problems and chessboard coloring.

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