445. At Lunch. The clerks of the firm "Pilkings and Popinjay" decided that they would sit three at a time at the same table every day until any 3 people would be forced to sit at this table again. The same number of clerks of the firm "Redson, Robson and Ross" decided to do the same, but with 4 people at a time. When they started to implement their plan, it turned out that the clerks of the second firm could continue rearranging themselves exactly three times longer than their neighbors.

What is the smallest number of clerks that could have served in each of the two firms?
445. If Pilkins had 11 clerks and Redson 12, they could have sat at a table in 165 and 495 ways respectively, which would have been the solution to the problem. However, we know that both firms had an equal number of clerks. Therefore, the answer is 15 clerks, sitting three at a time for 455 days, and 15 clerks, sitting four at a time for 1365 days.