How many solutions does the equation

$$
15 x+6 y+10 z=1973
$$

have in integers that satisfy the following inequalities:

$$
x \geq 13, \quad y \geq-4, \quad z>-6
$$
It is appropriate to introduce new unknowns whose smallest allowed value is the same number, for example, 1:

$$
x-12=x^{\prime} \geq 1, \quad y+5=y^{\prime} \geq 1, \quad z+6=z^{\prime} \geq 1
$$

where $x, y, z$ are integers. With these, (1) becomes:

$$
15 x^{\prime}+6 y^{\prime}+10 z^{\prime}=1883
$$

The second and third terms on the left side are divisible by 2, so based on the oddness of the right side, $15 x^{\prime}$, and thus $x^{\prime}$, must also be odd. Therefore, in every solution of (4),

$$
x^{\prime}=2 x^{\prime \prime}-1(\geq 1), \quad \text { where } \quad x^{\prime \prime} \geq 1, \text { is an integer }
$$

we introduce this as another unknown.

Similarly, from the fact that the first two terms of (4) are divisible by 3, and the right side is of the form $3k-1$, it follows that $10 z^{\prime}$ and $10 z^{\prime}-9 z^{\prime}=z^{\prime}$ also give a remainder of $-1$ when divided by 3. Considering (3), we get:

$$
z^{\prime}=3 z^{\prime \prime}-1, \quad \text { where } \quad z^{\prime \prime} \geq 1, \text { is an integer. }
$$

Finally, $15 x^{\prime}+10 z^{\prime}$ is divisible by 5, and 1883 can be written in the form $5k-2$, so the same must hold for $6 y^{\prime}$ and thus for $y^{\prime}$:

$$
y^{\prime}=5 y^{\prime \prime}-2, \quad \text { where } \quad y^{\prime \prime} \geq 1, \text { is an integer. }
$$

With our new unknowns, (4) becomes:

$$
x^{\prime \prime}+y^{\prime \prime}+z^{\prime \prime}=64, \quad x^{\prime \prime}, y^{\prime \prime}, z^{\prime \prime} \geq 1
$$

integers. Each solution of this equation provides a solution to the following problem: a stick of length $64 \mathrm{~cm}$ is divided into cm segments by division points; it must be divided into 3 parts at 2 division points such that the part containing the (red-marked) starting point of the stick goes to the I. person, the part containing the (blue-marked) end point of the stick goes to the III. person, and the middle part goes to the II. person. Conversely, every solution to the stick division problem is also a solution to (4), and thus to (1) under the appropriate conditions. - This completes the solution.

Remark. The result of 1953 can be interpreted as a year, and the small game with the unusual (2) inequalities to get the "current" year 1973 on the right side of (1) is understandable. It is not possible to get closer to 1973 with a binomial coefficient of the form $\binom{n}{2}$.