Problem 1228

To solve this problem, we need to ensure that every picket is painted exactly once by Harold, Tanya, and Ulysses. We will analyze the conditions under which this happens and find the sum of all paintable integers $100h + 10t + u$.

1. Analyzing the conditions:
- Harold starts with the first picket and paints every $h$-th picket.
- Tanya starts with the second picket and paints every $t$-th picket.
- Ulysses starts with the third picket and paints every $u$-th picket.

We need to ensure that every picket is painted exactly once. This means that the sequences generated by $h$, $t$, and $u$ must cover all positive integers without overlapping.

2. Case Analysis:
- **Case 1: $h$ paints the 4th picket.**
- If Harold paints the 4th picket, then $h = 3$.
- **Sub-case 1: $t$ paints the 5th picket.**
- If Tanya paints the 5th picket, then $t = 3$ and $u = 3$. This gives the paintable integer $333$.
- **Sub-case 2: $u$ paints the 5th picket.**
- If Ulysses paints the 5th picket, then $u = 2$. However, this would mean $(h, u) = 1$, which does not work because $h$ and $u$ must be relatively prime.

- **Case 2: $t$ paints the 4th picket.**
- If Tanya paints the 4th picket, then $t = 2$.
- **Sub-case 1: $h$ paints the 5th picket.**
- If Harold paints the 5th picket, then $h = 4$ and $u = 2$. This gives the paintable integer $424$.
- **Sub-case 2: $u$ paints the 5th picket.**
- If Ulysses paints the 5th picket, then $u = 2$. However, this would mean $h = 0$, which does not work because $h$ must be a positive integer.

- **Case 3: $u$ paints the 4th picket.**
- If Ulysses paints the 4th picket, then $u = 1$. However, this would mean $u$ paints everything, which does not work because $u$ must be greater than 1.

3. Summing the paintable integers:
- From the valid sub-cases, we have the paintable integers $333$ and $424$.
- Summing these integers: $333 + 424 = 757$.