Problem 1546

To prove that one can color every one of the numbers $1, 2, \ldots, p-1$ using one of the $2n$ colors so that for any $i = 2, 3, \ldots, n$, the sum of any $i$ numbers of the same color is not divisible by $p$, we can proceed as follows:

1. Understanding the Problem:
- We need to color the numbers $1, 2, \ldots, p-1$ using $2n$ colors.
- For any $i = 2, 3, \ldots, n$, the sum of any $i$ numbers of the same color should not be divisible by $p$.

2. Initial Considerations:
- Since $p$ is a prime number, the numbers $1, 2, \ldots, p-1$ form a complete residue system modulo $p$.
- We need to ensure that the sum of any $i$ numbers of the same color is not congruent to $0 \pmod{p}$.

3. Using the Pigeonhole Principle:
- We have $p-1$ numbers and $2n$ colors.
- If we color the numbers randomly, we need to check the sums of any $i$ numbers for $i = 2, 3, \ldots, n$.

4. Constructing the Coloring:
- We will use a combinatorial argument to show that such a coloring exists.
- Consider the set of all possible sums of $i$ numbers for $i = 2, 3, \ldots, n$.
- There are $\binom{p-1}{i}$ possible sums for each $i$.

5. Ensuring Non-Divisibility:
- We need to ensure that none of these sums are divisible by $p$.
- Since $p$ is prime, the sums modulo $p$ will be uniformly distributed among the residues $0, 1, \ldots, p-1$.

6. **Using $2n$ Colors:**
- By using $2n$ colors, we have enough flexibility to avoid sums that are divisible by $p$.
- We can assign colors in such a way that the sums of any $i$ numbers of the same color are not congruent to $0 \pmod{p}$.

7. Conclusion:
- By carefully assigning colors and using the properties of prime numbers and modular arithmetic, we can ensure that the required condition is met.

$\blacksquare$