Problem 1025

Solution:

The answer is 1706. Indeed, the squares $(x, y)$ on which there is a checker can be divided into two groups:

a. $x$ a multiple of 2 and $y$ a multiple of 3;
b. $x$ a multiple of 3 and $y$ a multiple of 2.

Since there are on the chessboard $\left[\frac{75}{2}\right]$ columns with abscissa a multiple of 2 and $\left[\frac{75}{3}\right]$ rows with ordinate a multiple of 3, the number of squares 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 $x$, that is the greatest integer less than or equal to $x$).

By symmetry (since the chessboard is square, it suffices to rotate it by $90^\circ$) the number of squares of the second type is the same. In this way, however, we have counted twice the squares that belong to both groups, for example $(6,6)$. By the way the two groups are defined, these squares are all and only those in which the abscissa and the ordinate are both multiples of 6, whose number 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 squares that have been counted twice. To sum up, we have

Squares with a checker $=$ Squares in the first group + Squares in the second group $-$ Squares in both groups $=$
$$= 925 + 925 - 144 = 1706$$