If we denote with $x$ the number of boards of $8 \, \text{cm}$ width, with $y$ those of $10 \, \text{cm}$, and with $z$ those of $12 \, \text{cm}$ width, then $8x + 10y + 12z = 90$ or, if we divide both sides of the equation by 2, $4x + 5y + 6z = 45$.

This is an equation with three unknowns. Such an equation initially has infinitely many solutions. However, the number of solutions is significantly restricted by the conditions of the problem.

1. Only positive solutions are possible, as a negative number of boards makes no sense. This means that the following must hold:

$$
0 \leq 4x \leq 45, \, 0 \leq 5y \leq 45, \, 0 \leq 6z \leq 45 \rightarrow 0 \leq x \leq 11, \, 0 \leq y \leq 9, \, 0 \leq z \leq 7.5
$$

2. Only integer solutions are possible, as no board should be cut lengthwise. This diophantine problem is solved as follows:

From $4x + 5y + 6z = 45$, by subtracting $5y$ from both sides of the equation, we get $4x + 6z = 45 - 5y$. The left side of the equation is divisible by 2, so the right side must also be divisible by 2. Since 45 is an odd number, $5y$ and thus $y$ must also be an odd number. Due to condition 1, this results in five possibilities for $y$: $y=1, y=3, y=5, y=7, y=9$.

Substituting these five values into the equation one by one, we obtain five equations, each containing only two unknowns:

1. $y=1: \quad 4x + 6z = 45 - 5 \rightarrow 2x + 3z = 20$
2. $y=3: \quad 4x + 6z = 45 - 15 \rightarrow 2x + 3z = 15$
3. $y=5: \quad 4x + 6z = 45 - 25 \rightarrow 2x + 3z = 10$
4. $y=7: \quad 4x + 6z = 45 - 35 \rightarrow 2x + 3z = 5$
5. $y=9: \quad 4x + 6z = 45 - 45 \rightarrow 2x + 3z = 0$

These five equations are treated in the same way. However, in the case of $5 (y=9)$, it is immediately clear that $x=0$ and $z=0$ follows, i.e., one possibility is to fill the door opening with nine boards of $10 \, \text{cm}$ width. Thus, $x_1 = 0, y_1 = 9, z_1 = 0$.

In the case of $1 (y=1)$, it follows from $2x + 3z = 20$ that $2x = 20 - 3z$. Since $2x$ is divisible by 2, $20 - 3z$ and thus $z$ must also be divisible by 2. Therefore, the following four possibilities apply to $z$: $z_2 = 0, z_3 = 2, z_4 = 4, z_5 = 6$. (For $z \geq 8$, $x < 0$ would follow, contradicting condition 1); for $x$, we get: $x_2 = 10, x_3 = 7, x_4 = 4, x_5 = 1$.

In cases 2, 3, and 4, the same approach is followed, with the results:

1. $y=3: \quad z_6 = 1, z_7 = 3, z_8 = 5$ and $x_6 = 6, x_7 = 3, x_8 = 0$
2. $y=5: \quad z_9 = 0, z_{10} = 2$ and $x_9 = 5, x_{10} = 2$
3. $y=7: \quad z_{11} = 1$ and $x_{11} = 1$

Thus, a total of eleven solutions to the problem have been found. From the solution process, it is clear that there can be no further solutions.
## Task 11/61

A door opening of $90 \mathrm{~cm}$ in width is to be nailed shut with boards. Available are boards of suitable length and widths of $8 \mathrm{~cm}$, $10 \mathrm{~cm}$, and $12 \mathrm{~cm}$.

What are the possible combinations if no board is to be cut lengthwise?