13. On a $75 \times 75$ chessboard, the rows and columns are numbered from 1 to 75. Chiara wants to place a pawn on all and only the squares that have one coordinate even and the other a multiple of 3. How many pawns will she place in total on the chessboard?
13. The answer is 1706. Indeed, the cells $(x, y)$ where there is a pawn can be divided into two groups:

(a) $x$ is a multiple of 2 and $y$ is a multiple of 3;

(b) $x$ is a multiple of 3 and $y$ is a multiple of 2.

Since there are $\left[\frac{75}{2}\right]$ columns with an abscissa that is a multiple of 2 and $\left[\frac{75}{3}\right]$ rows with an ordinate that is a multiple of 3, the number of cells of the first type is

$$
\left[\frac{75}{2}\right] \cdot\left[\frac{75}{3}\right]=37 \cdot 25=925
$$

(here $[x]$ denotes the integer part of $x$, i.e., the largest integer less than or equal to $x$).

By symmetry (since the board is square, it is enough to rotate it by $90^{\circ}$), the number of cells of the second type is the same. However, in this way, we have counted twice the cells that belong to both groups, for example, $(6,6)$. According to how the two groups are defined, such cells are all and only those where the abscissa and ordinate are multiples of 6, the number of which is

$$
\left[\frac{75}{6}\right] \cdot\left[\frac{75}{6}\right]=12 \cdot 12=144
$$

We must therefore subtract from the total the number of these cells that have been counted twice. In summary, we have

Cells with a pawn $=$ Cells in the first group+

+ Cells in the second group - Cells in both groups $=$

$$
=925+925-144=1706
$$