10.5. Uncle Chernomor assigns 9 or 10 of his thirty-three bogatyrs (knights) to duty each evening. What is the smallest number of days after which it can happen that all bogatyrs have been on duty the same number of times?
# 10.5. 7.

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Let \( m \) and \( n \) be the number of days when 9 and 10 heroes were on duty, respectively. Let \( k \) be the number of days each hero was on duty. Then \( 9m + 10n = 33k \). Note that from this, \( n \) must be divisible by 3.

Firstly, the equation \( 9m + 10n = 33 \) has no solutions. Indeed, \( n \) cannot be greater than 3, so \( n \) is either 0 or 3, but \( 33 - 10n \) is not divisible by 9.

Secondly, the equation \( 9m + 10n = 66 \) has a unique solution in non-negative integers \( m = 3, n = 4 \). This is easy to verify by substituting \( n \) with 0, 3, 6. Note that \( m + n = 7 \).

Furthermore, if \( k \geq 3 \), then from the inequalities \( 10(m + n) \geq 9m + 10n = 33k \geq 99 \) it follows that \( m + n > 8 \).

It remains to provide an example of a duty schedule for 7 days. Assign the heroes numbers from 1 to 33. First day: 1-9; second day: 10-18; third day: 19-27; fourth day: 27-33 and 1-3; in the remaining days, 10 heroes in order.