Problem 1283

1. Understanding the Problem:
We need to find the sum $S$ of the products of the non-zero digits of all numbers from 1 to 999. For a number $n$, $p(n)$ is defined as the product of its non-zero digits.

2. Breaking Down the Problem:
We can consider the numbers from 1 to 999 as three-digit numbers by padding with leading zeros if necessary. For example, 5 can be considered as 005.

3. **Calculating $p(n)$ for Each Digit:**
- For a single-digit number $n$, $p(n) = n$.
- For a two-digit number $n = 10a + b$, $p(n) = a \cdot b$ if $a \neq 0$ and $b \neq 0$.
- For a three-digit number $n = 100a + 10b + c$, $p(n) = a \cdot b \cdot c$ if $a, b, c \neq 0$.

4. **Summing $p(n)$ for All Numbers from 1 to 999:
We need to consider the contribution of each digit position (hundreds, tens, and units) separately.

5. Contribution of Each Digit Position:**
- Each digit from 1 to 9 appears in each position (hundreds, tens, units) exactly 100 times in the range from 1 to 999.
- The sum of the non-zero digits from 1 to 9 is $1 + 2 + 3 + \cdots + 9 = 45$.

6. Calculating the Total Sum:
- For each digit position, the sum of the products of the non-zero digits is $45$ (since each digit from 1 to 9 appears 100 times).
- Therefore, the total sum for each position is $45 \times 100 = 4500$.

7. Combining the Contributions:
- Since there are three positions (hundreds, tens, units), the total sum $S$ is $4500 + 4500 + 4500 = 13500$.

8. Correcting the Calculation:
- We need to consider the product of the non-zero digits for each number. For example, for 105, $p(105) = 1 \times 5 = 5$.
- The correct approach is to consider the product of the sums of the non-zero digits for each position.

9. Revisiting the Calculation:
- The sum of the products of the non-zero digits for each position is $(1 + 1 + 2 + 3 + \cdots + 9) = 46$.
- Therefore, the total sum $S$ is $46^3$.

10. Final Calculation:
$$S = 46^3 = 46 \times 46 \times 46 = 97336$$

11. Finding the Largest Prime Factor:
- We need to find the prime factors of 97336.
- The prime factorization of 97336 is $2^3 \times 11567$.
- The largest prime factor is 11567.

The final answer is $\boxed{11567}$.