Sílvia is going to fill her 10 water jugs at a source that has three taps. One of the jugs takes one minute to fill, another two minutes, another three minutes, and so on. How should Sílvia distribute the 10 jugs among the three taps to spend the least amount of time possible? What is this time?
To simplify, we number the 10 water dispensers according to the respective times they take to fill up, from 1 to 10.

Solution 1: One idea is to use the "remaining time" of one water dispenser to fill another, filling two others simultaneously. The following figures illustrate the solution. In Figure I, the 3 taps take 10 minutes to fill the dispensers 10,

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-072.jpg?height=437&width=460&top_left_y=547&top_left_x=341)

Figure II: 9 min

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-072.jpg?height=322&width=366&top_left_y=650&top_left_x=1439)

In Figure I, the 3 taps take 10 minutes to fill the dispensers 10, 9, 8, 2, and 1, and in Figure II, the 3 taps take another 9 minutes to fill the dispensers 7, 6, 5, 4, and 3. Therefore, the total time spent is 19 minutes.

Solution 2: If we had only one tap, the time spent to fill the 10 dispensers would be $1+2+\cdots+9+10=55$ minutes. Since we have three taps and $55=3 \times 18+1$, at least one tap will take 19 minutes, and the other two will take 18 minutes each. The following table shows one way to do the work in 19 minutes.

| Tap 1 | 10 | 9 |  |  |
| :---: | :---: | :---: | :---: | :---: |
| Tap 2 | 8 | 5 | 3 | 2 |
| Tap 3 | 7 | 6 | 4 | 1 |