6. There are three vessels with volumes of 3 liters, 4 liters, and 5 liters, without any markings, a water tap, a sink, and 3 liters of syrup in the smallest vessel. Can you, using pourings, obtain 6 liters of a water-syrup mixture such that the amount of water is equal to the amount of syrup in each vessel?
Solution.

For example, as follows (see the table below, c - syrup, w - water, f - final mixture).

|  | ![](https://cdn.mathpix.com/cropped/2024_05_06_b6522f95bc9413ab322cg-5.jpg?height=128&width=365&top_left_y=628&top_left_x=909) | 4-liter  container | 5-liter  container |
| :---: | :---: | :---: | :---: |
| Pour the syrup into the 5-liter container, and using the 3- and 4-liter containers, get 2 liters of water in the 4-liter container. | $3 \mathrm{c}$ | 0 | 0 |
|  | 0 | 0 | $3 \mathrm{c}$ |
|  | Zw | 0 | $3 \mathrm{c}$ |
|  | 0 | Zw | $3 \mathrm{c}$ |
|  | Zw | Zw | $3 \mathrm{c}$ |
|  | 2B | 4B | $3 \mathrm{c}$ |
|  | 2B | 0 | $3 \mathrm{c}$ |
|  | 0 | 2B | $3 \mathrm{c}$ |
| Then fill the 4-liter container with syrup to get 4 liters of the required mixture. | 0 | $2 \mathrm{~B}+2 \mathrm{c}=4$ f | $1 \mathrm{c}$ |
| ![](https://cdn.mathpix.com/cropped/2024_05_06_b6522f95bc9413ab322cg-5.jpg?height=293&width=711&top_left_y=1374&top_left_x=199) | Zi | 1f | $1 \mathrm{c}$ |
|  | Zi | $1 f+1 c$ | 0 |
|  | 0 | $1 f+1 c$ | Zi |
|  | $1 f+1 c$ | 0 | Zi |
|  | 1 f $+1 c+1$ w $=3$ f | 0 | Zi |

## Grading Criteria.

- Correct algorithm - 7 points.
- Less than 6 liters of the required mixture obtained (i.e., the required mixture is obtained in one container, but the ratio is not maintained in the other) — no more than 1 point.