We want to move a piece from the top left corner to the bottom right corner of a $5 \times 5$ chessboard, where the piece can only move one step to the right or one step down. We want to mark one forbidden square on the chessboard, which the piece cannot step on. Which square should this be, so that we can reach the bottom right corner from the top left corner in the fewest possible ways, but still be able to reach it?
If we want to block the path from the top left corner to the bottom right corner, then these two squares cannot obviously be covered. Suppose that we can reach a certain square from the top left corner in $i$ ways, and from there to the bottom right in $j$ ways. Then covering this square deprives the pawn of $i \cdot j$ paths. We need to find when this product is the largest. First, let's see how many ways we can reach each square. The top row and the left column can only be reached in one way. The pawn can only move to any other square from above or from the left.

| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :---: | :---: | :---: |
| 1 | 2 | 3 | 4 | 5 |
| 1 | 3 | 6 | 10 | 15 |
| 1 | 4 | 10 | 20 | 35 |
| 1 | 5 | 15 | 35 | 70 |

If we know that the square above a selected square can be reached in $p$ ways, and the square to the left in $q$ ways from the top left corner, then the selected square can be reached in $(p+q)$ ways. Based on this, we can sequentially write the number of ways to reach each square on the chessboard. Now we also need to examine how many ways we can reach the bottom right corner from each square. The number of these paths is obviously the same as the number of paths from the top left corner to the square that is centrally symmetric to it. Therefore, we only need to read from the diagram which pair of centrally symmetric squares has the largest product of the numbers in them. It is easily verifiable that this is the case for the pair of squares containing the numbers 2 and 20. Thus, by blocking one of these, the pawn can reach the bottom right corner from the top left in the fewest possible ways.

Péter Regös (Miskolc, Földes F. Gymnasium, 1st year student)