Solution. Note that there are 32 white cells, and each part contains an integer number of white cells, so the answer must be a divisor of 32.

Obviously, it cannot be 1. The answers $2, 4, 8, 16$, and 32 are possible. To achieve them,

![](https://cdn.mathpix.com/cropped/2024_05_06_b5ab07e644997c69f8fdg-1.jpg?height=504&width=395&top_left_y=1225&top_left_x=2439)
we need to cut the board into rectangles $8 \times 8, 8 \times 4, 8 \times 2, 4 \times 2, 2 \times 2$, and then divide each of them into two equal parts. Constructions for two rectangles are shown in the figure; the others are similar.

Criteria for determining winners and prizewinners of the final stage of the Youth Mathematical School Olympiad, 2015-16.

Each problem is worth 1 point.

For 4th grade:

A winner is a participant who solved 6 problems.

A prizewinner is a participant who solved at least 4 problems.

For 5th grade:

A winner is a participant who solved at least 6 problems.

A prizewinner is a participant who solved at least 4 problems.

For 6th grade:

A winner is a participant who solved 7 problems.

A prizewinner is a participant who solved at least 5 problems.

For 7th grade:

A winner is a participant who solved 7 problems.

A prizewinner is a participant who solved at least 5 problems.

For 8th grade:

A winner is a participant who solved 6 problems.

A prizewinner is a participant who solved at least 4 problems.

For 9th grade:

A winner is a participant who solved 8 problems.

A prizewinner is a participant who solved at least 6 problems.

For 10th grade:

A winner is a participant who solved 9 problems.

A prizewinner is a participant who solved at least 5 problems.

For 11th grade:

A winner is a participant who solved 7 problems.

A prizewinner is a participant who solved at least 5 problems.
7. A chessboard $(8 \times 8)$ was cut into several equal parts in such a way that all white cells remained uncut, while each black cell was cut. How many parts could have been obtained?