Question

На поле 2007на2007 стоят 2007 не бьющих друг друга ладьей. Каждую из них сдвинули на соседнюю клетку. Докажите, что теперь какие то две ладьи бьют друг друга

Answer 1

Чтобы на поле разместилось столько же ладьей, сколько клеток в одной стороне, они должны быть выставлены точно по диагонали из одной самой крайней клетки до другой. Если сместить каждую из них на одно поле, то самая последняя ладья не поместится в получившейся диагонали и ее тогда придется подвинуть в другую сторону, где ее уже бьет стоящая на той же линии другая ладья. надеюсь, я достаточно понятно объяснил;)

Answer 2

Problem Analysis

The problem states that there are 2007 rooks (also known as "lad'ya" or "castle") placed on a 2007 by 2007 chessboard. Each rook is moved to an adjacent square. The task is to prove that there are now two rooks that are attacking each other.

Solution

To solve this problem, we can use the pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon.

In this case, we have 2007 rooks and 2007 squares on the chessboard. Since each rook is moved to an adjacent square, each rook will occupy a different square after the move. Therefore, we have 2007 rooks and 2007 squares, which means that each rook occupies a unique square.

Now, let's consider the possible positions of the rooks after the move. Each rook can move horizontally or vertically to an adjacent square. Since there are 2007 rooks, there are a total of 2007 possible positions for the rooks after the move.

However, there are only 2007 squares on the chessboard. Therefore, by the pigeonhole principle, at least one square must contain more than one rook. This means that there are two rooks that are attacking each other.

Therefore, we have proved that after moving the 2007 rooks to adjacent squares, there will always be two rooks that are attacking each other.

Note: The search results provided did not contain any relevant information for this specific problem. However, the solution provided is based on the pigeonhole principle, which is a well-known mathematical principle used to solve similar problems.