Problem 1009

It is only possible to fill a large rectangle with smaller rectangles if the sides of the smaller rectangles are parallel to the corresponding sides of the large rectangle. Let's place the "large" rectangle on a chessboard as shown in the diagram, where the sides of the squares are $1 / 2$ in length. It is easy to see that each of the smaller rectangles in the division covers an equal area of white and black parts of the chessboard. (This follows from the fact that one side of each small rectangle is an integer.) Therefore, the large rectangle also covers an equal area of white and black parts. Thus, we prove the statement of the problem if we show that this is only possible when one of the sides of the large rectangle is an integer.

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Suppose this is not the case. Cut out from the lower right corner of the rectangle a rectangle whose sides are the amounts by which the two sides exceed the nearest integer. The remaining (concave) hexagon has an equal area of white and black parts (since it splits into two rectangles, each with one integer side), so this is also true for the cut-out part. However, the sides of the cut-out rectangle fall between 0 and 1, so the black part in it must have a larger area. The contradiction proves our statement.

Remarks. 1. A special case of the problem statement is the following. A rectangle can be tiled with $1 \times k$ rectangles if and only if both sides are integers and one of them is divisible by $k$.

2. Both the statement and the proof can be easily generalized to three-dimensional rectangular prisms.