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?
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.