456. Every evening, Uncle Chernomor appoints 9 or 10 out of 33 bogatyrs for duty, at his discretion. What is the smallest number of days after which it can happen that each of the bogatyrs has been on duty the same number of times?
456. Answer: 7 days. Let $k$ be the number of days when 9 bogatyrs were on duty, and $l$ be the number of days when 10 bogatyrs were on duty, with each of them being on duty $m$ times. Then $9 k + 10 l = 33 m$. For $m=1$ there are no solutions, but for $m=2$ we have $k=4$ and $l=3$.