10.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with? (N. Agakhanov)
Answer: Up to 8 zeros.

Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the sum of our numbers is not less than $9 \cdot 10^{9}$. This means that one of them is not less than $10^{9}$, which is impossible.

It remains to show how to form numbers whose sum ends with eight zeros. For example, we can take eight numbers equal to 987654321, and one number 198765432. Their sum is $81 \cdot 10^{8}$.