There are 10 compartments in a passenger train which carries on an average 20 passengers per compartment. If at least 12 passengers were sitting in each compartment and no compartment has an equal number of passengers, then what is the maximum number of passengers that can be accommodated in any compartment?

10 compartments, average 20 passengers = total 200 passengers. Each compartment has at least 12 passengers, all different. To maximize one compartment, minimize the others: 12+13+14+15+16+17+18+19+20=146. Remaining for 10th compartment: 200-146=54. But 54 conflicts with existing numbers. The correct maximization is: 12+13+14+15+16+17+18+19+20+21 = 165, which doesn't average to 20. The actual solution: minimum sum for 9 compartments = 12+13+14+15+16+17+18+19+20 = 144. Then 200-144=56.