nine (not necessarily distinct) nine-digit numbers have been formed; each digit has been used in each number exactly once. What is the maximum number of zeros that the sum of these nine numbers can end with

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Evaluation. Let the sum end with nine 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 that is divisible by 9 and ends with nine zeros is $9 \cdot 10^{9}$, so the sum of our numbers is no less than $9 \cdot 10^{9}$. This means that one of them is no less than $10^{9}$. Contradiction.

Example with eight zeros: $8 \cdot 987654321 + 198765432 = 81 \cdot 10^{8}$.

## Answer

To 8 zeros.