10.3. What is the maximum number of digits that a natural number can have, where all digits are different, and it is divisible by each of its digits?
Answer: 7 digits.

Evaluation. There are 10 digits in total. The number cannot contain the digit 0, so there are no more than 9. If all 9, then the digit 5 must be at the end of the number (divisibility rule for 5), but in this case, the number must also be divisible by 2. Therefore, the digit 5 is not in this number. If the number lacks only 0 and 5, then it contains 9, and the sum of the digits of this number will be 40, which contradicts divisibility by 9. Thus, there are no more than seven digits.

An example of such a number is 9176328. Verification. The three-digit number formed by the last three digits is 328, which is divisible by 8, meaning the given number is divisible by its digits 2 and 8. The sum of the digits of this number is 36, so it is divisible by 3 and 9, and also by 6. It is easy to see that it is divisible by 7: $9176328: 7=1310904$.

Remarks. An example of a seven-digit number with verification is provided - 3 points. It is proven that there are no more than seven digits, but an example of such a number is not provided - 3 points. If both are present, 7 points.