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 many tokens Ana can place on a 2020 × 2020 board following the rules: no two tokens can share a row, column, or diagonal. This means that the tokens are placed such that they have no common points with other tokens.

The strategy to solve this problem is to realize that Ana needs to place the tokens in a way that maximizes their number while following the given restrictions. Let's analyze the situation:

1. **Placement Strategy:**
   - The best way to place the tokens so that they do not share any row, column, or diagonal is to use a pattern similar to a checkerboard.
   - In this pattern, the board is divided into regions where no two tokens can attack each other. This requires placing tokens such that they are spaced out.

2. **General Formula:**
   - The maximum number of tokens that can be placed on an \( n \times n \) board, given these constraints, corresponds to placing tokens on either all even-indexed or odd-indexed squares of each row and column. 
   - Specifically, if \( n \) is even, the formula for the maximum number of non-attacking tokens is \( \frac{n^2}{2} \).

3. **Calculation for a 2020 × 2020 Board:**
   - For a 2020 × 2020 board:
     \[
     n = 2020
     \]
     \[
     \text{Maximum number of tokens} = \frac{2020^2}{2} = \frac{4080400}{2} = 2040200
     \]

The correct answer, therefore, is \( 2020^2 \), which corresponds to option **(E) 2020^2**.

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

Here are some related questions to expand your understanding:
1. How would the solution change if Ana could only place tokens in one color on a checkerboard?
2. Can you describe the placement strategy for an \( n \times n \) board when \( n \) is odd?
3. How does the placement strategy change if the tokens can attack diagonally?
4. What is the maximum number of tokens on a 4 × 4 board using this method?
5. How many tokens can be placed on a 9 × 9 board with this strategy?
6. How can this method be adapted for a non-square board, such as 2020 × 2021?
7. Can this approach be generalized to higher dimensions?
8. What is the difference in the strategy if diagonal attacks are allowed?

**Tip:** When dealing with board problems, breaking down the problem into patterns like checkerboards often reveals optimal solutions for arranging non-attacking pieces.

Discover how to maximize the number of tokens on a 2020 x 2020 board using a non-attacking placement strategy. Follow step-by-step solutions and learn key combinatorial concepts. Suitable for Grades 9-12.

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