## Task 1 - 240611

For the Pioneer Birthday, the most diligent waste collectors are to be honored. For this purpose, the Pioneer leader wants to buy books for $6 \mathrm{M}$ and $4 \mathrm{M}$ each, at least one of each type, but no other types. In total, she wants to spend $30 \mathrm{M}$ on these books.

List all the possibilities for the quantities of books of the two types that can be chosen to meet these conditions!
When $x$ books at $6 \mathrm{M}$ each and $y$ books at $4 \mathrm{M}$ each are bought, the equation $6 x + 4 y = 30$ (1) holds.

Given $y \geq 1$, it follows that $6 x \leq 26$, so $x < 5$. Therefore, and because $x \geq 1$, there are only the following possibilities for $x$ in the table below. Those values of $x$ for which the number $30 - 6 x$ is not divisible by 4 are excluded, since from (1) it follows that $4 y = 30 - 6 x$ must hold.

For the remaining values of $x$, the corresponding values for $y$ are derived from this equation.

| $x$ | $6 x$ | $30 - 6 x = 4 y$ | $y$ |
| :---: | :---: | :---: | :---: |
| 1 | 6 | 24 | 6 |
| 2 | 12 | 18 | - |
| 3 | 18 | 12 | 3 |
| 4 | 24 | 6 | - |

Therefore, only the following quantities can meet the conditions of the problem:

Either 1 book at $6 \mathrm{M}$ and 6 books at $4 \mathrm{M}$, or 3 books at $6 \mathrm{M}$ and 3 books at $4 \mathrm{M}$ are bought.

Both quantity specifications satisfy the conditions (1) and $x \geq 1, y \geq 1$. Therefore, all the required possibilities are listed here.