Maths Olympiad Prep

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Combinatorics Difficulty 5.1 AIME, harder Find the answer

Example 3 - What is the maximum number of knights that can be placed on an 8×88 \times 8 chessboard so that no two knights attack each other (assuming there are enough knights)?

A number or a short expression. Spacing and $ signs are ignored.

Solution

We will alternately color the chessboard in black and white, so there will be 32 black squares and 32 white squares. According to the knight's move (see Figure 1), a knight on a black square can only capture a knight on a white square. Therefore, placing knights on all black squares means they will not capture each other. This means we can place 32 knights, and they will not capture each other. Now, we need to prove that placing 33 knights will inevitably result in some being captured.

In fact, dividing the chessboard

into 8 smaller 2×42 \times 4
boards (as shown in Figure 6),
at least one of these smaller
boards will have to contain
5 knights. The possible
placements are: either
one row has 1 knight and the other has 4; or one row has 2 knights and the other has 3. Clearly, both of these placements will inevitably result in knights capturing each other.
Therefore, the maximum number of knights that can be placed so that they do not capture each other is 32.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.