Task 10.4. In the central cell of a $21 \times 21$ board, there is a chip. In one move, the chip can be moved to an adjacent cell by side. Alina made 10 moves. How many cells can the chip end up in?
Answer: 121.

Solution. We will paint the entire board in a checkerboard pattern so that the central cell of the board is black. With each move of the chip to an adjacent cell, the color of the cell on which the chip is standing will change. After an odd number of moves, the chip will always be on a white cell, and after an even number of moves - on a black cell. Therefore, after 10 moves, the chip will definitely be on a black cell.

We will show that to all black cells that can be reached in no more than 10 moves, it is possible to reach them exactly in 10 moves. Consider an arbitrary black cell $A$ that can be reached in fewer than 10 moves from the center. Since by the time the chip reaches cell $A$, an even number of moves, less than 10, has been made, one can simply move the chip to an adjacent cell and back until exactly 10 moves have been made. Therefore, we need to count the number of black cells that can be reached in no more than 10 moves.

In 0 moves, you can only reach the starting cell, in 2 moves you can reach the starting cell and $4 \cdot 2=8$ new cells, in 4 moves you can reach the cells that have already been reached and $4 \cdot 4=16$ new cells, and so on. We get that the number of cells that can be reached in no more than 10 steps is $1+8+16+24+\ldots+40=$ $1+8(1+2+3+4+5)=121$

Remark. The centers of the cells that can be reached in $k \leqslant 10$ steps form the vertices of a square grid rotated by $45^{\circ}$ (Fig. 6). From this, it is clear that the total number of such cells is $11^{2}$.