The product of the digits of any multi-digit number is always less than the number itself. If we calculate the product of the digits of a given multi-digit number, then the product of the digits of this product, and then the product of the digits of the new product, etc., we will inevitably reach a single-digit number after a certain number of steps. This number of steps is called the persistence of the number. For example, the number 723 has a persistence of 2, because $7 \cdot 2 \cdot 3=42$ (1st step) and $4 \cdot 2=8$ (2nd step).

1. Find the largest odd number with distinct digits and a persistence of 1.
2. Find the largest even number with distinct non-zero digits and a persistence of 1.
3. Find the smallest natural number with a persistence of 3.

(S. Bednářová)
1. The problem statement does not specify that $\mathrm{v}$ cannot use zero in this case. If one of the digits is zero, it means that the product in the first step is also zero, and thus the persistence is 1. Therefore, it is sufficient to construct the largest odd number with distinct digits; this is 9876543201.

2. This time, we cannot use zero. This means that the digit product of the number we are looking for must be a single-digit number, while we try to obtain as many distinct factors as possible (the number of factors then determines the number of digits in this number, i.e., the more factors, the higher the number). Let's consider all possible factorizations of single-digit numbers into products of natural numbers.

Since we are looking for an even number, we need at least one factor of the digit product to be an even number. This means that the digit product is also an even number, so we can limit our search for factorizations to the numbers $2, 4, 6$, and 8. Furthermore, we can focus only on factorizations that include 1 as a factor. The corresponding numbers are always one order of magnitude higher than the numbers corresponding to factorizations without 1.

- $2=1 \cdot 2$, options: 12,
- $4=1 \cdot 4$, options: 14,
- $4=1 \cdot 2 \cdot 2$, not possible (identical factors),
- $6=1 \cdot 6$, options: 16,
- $6=1 \cdot 2 \cdot 3$, options: 132, 312,
- $8=1 \cdot 8$, options: 18,
- $8=1 \cdot 2 \cdot 4$, options: 124, 142, 214, 412.

Among the found options, the highest number is 412.

3. We can solve this task by sequentially checking multi-digit numbers starting from the smallest (i.e., 10) and determining their persistence. The first found number with a persistence of 3 is the desired number.

Two-digit numbers containing the digit 1 or 0 have a persistence of 1 because the corresponding digit product is at most 9. Similarly, two-digit numbers containing the digit 2 have a persistence of at most 2 because the corresponding digit product is at most 18. Based on these considerations, it is sufficient to start checking natural numbers from 33:

- $33, 3 \cdot 3=9$, persistence 1,
- $34, 3 \cdot 4=12, 1 \cdot 2=2$, persistence 2,
- $35, 3 \cdot 5=15, 1 \cdot 5=5$, persistence 2,
- $36, 3 \cdot 6=18, 1 \cdot 8=8$, persistence 2,
- $37, 3 \cdot 7=21, 2 \cdot 1=2$, persistence 2,
- $38, 3 \cdot 8=24, 2 \cdot 4=8$, persistence 2,
- $39, 3 \cdot 9=27, 2 \cdot 7=14, 1 \cdot 4=4$, persistence 3.

The smallest natural number with a persistence of 3 is 39.