We cut an $8 \times 8$ chessboard into $p$ rectangular pieces such that no single square is cut in half. Each such cutting must satisfy the following conditions:

(a) Each rectangle must contain an equal number of white and black squares.

(b) If $a_{i}$ denotes the number of white squares in the $i$-th rectangle, then the sequence $a_{1}<a_{2}<\ldots<a_{p}$ must hold.

Find the largest value of $p$ for which such a cutting exists.

Furthermore, provide all sequences of $a_{1}, a_{2}, \ldots, a_{p}$ corresponding to this value of $p$.
Since there are 32 white squares on a chessboard,

$$
a_{1}+a_{2}+\ldots+a_{p}=32
$$

It follows that $p$ can be at most 7, because according to condition $b$, if $p \geq 8$,

$$
a_{1}+a_{2}+\ldots+a_{p} \geq 1+2+\ldots+8=36>32
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=233&width=246&top_left_y=539&top_left_x=522)
a)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=233&width=241&top_left_y=539&top_left_x=801)
b)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=243&width=225&top_left_y=529&top_left_x=1080)
c)

![](https://cdn.mathpix.com/cropped/2024_05_02_d2753a75c03c296983f7g-1.jpg?height=247&width=241&top_left_y=527&top_left_x=1338)
d)

We need to examine whether $p=7$ is possible. After some trial and error, we can find a partition that meets the conditions of the problem, as shown in part a) of the figure. This answers the first part of the problem: the maximum value of $p$ is 7.

To generate all possible sequences for $p=7$, we need to determine which numbers can appear among $a_{1}, \ldots, a_{7}$. Any number from 1 to 10 can appear. 11 cannot, because a rectangle of 22 squares cannot be cut from an $8 \times 8$ square. Any number larger than 11 cannot appear either, because then the sum of the other six different natural numbers could be at most $32-12=20$, but $1+2+\ldots+6=21$.

At least one of the numbers 8, 9, 10 must appear, because the sum of the first 7 natural numbers is only 28. The numbers 8 and 10, or 9 and 10, cannot appear together, because the sum of the other five different terms would have to be 14 or 13, which is not possible because $1+2+3+4+5=15$. Depending on which of the numbers 8, 9, 10 appear, we have four possible cases:

a) Only the 8 appears: $1+2+\ldots+8=36$, and we need to remove one of these to get a sum of 32. This can only be achieved by removing the 4. The resulting sequence is: 1, 2, 3, 5, 6, $7,8$.

b) Only the 9 appears: Similarly, by removing the 5, we get the sequence 1, 2, 3, 4, 6, 7, 9.

c) Only the 10 appears: By removing the 6, we get the sequence $1,2,3,4,5,7,10$.

d) Both 8 and 9 appear. We need to remove two numbers from $1,2, \ldots, 8,9$ to reduce the sum by $45-32=13$, but we must keep the 8 and 9. This can only be achieved by removing the 6 and 7, resulting in the sequence: $1,2,3,4,5,8,9$.

Thus, at most the four sequences listed above can correspond to the value $p=7$. It remains to show that these sequences can indeed be obtained by partitioning. A partition for each sequence is shown in the figure.

Neumer Attila (Budapest, Fazekas M. Gyak. Gimn., I. o. t.)