There are four bulbs and in a room. glows for 2 minutes, then goes off for 1 minute and the process continues in this manner. Similarly, glows for 3 minutes and then goes off for 1 minute and the process continues. For , it is 4 minutes and 1 minute, while for , it is 5 minutes and 1 minute respectively for glowing and going off. A man enters the room and comes out of it as soon as he completes an hour. What is the maximum number of times that he may face complete darkness in the room?

The problem involves finding the intersection of 'off' periods for four bulbs with different cycles. Bulb A is off at minutes 3, 6, 9... (multiples of 3). Bulb B is off at minutes 4, 8, 12... (multiples of 4). Bulb C is off at minutes 5, 10, 15... (multiples of 5). Bulb D is off at minutes 6, 12, 18... (multiples of 6). Complete darkness occurs when all bulbs are off simultaneously. This requires a minute that is a common multiple of 3, 4, 5, and 6. The Least Common Multiple (LCM) of 3, 4, 5, and 6 is 60. In a 60-minute period, the only common multiple is 60. However, the man enters at an arbitrary time. If he enters at minute 0, he faces darkness at minute 60. If he enters just after minute 0, he might face darkness at minute 0 and minute 60, totaling 2 instances. Thus, the maximum number of times he may face complete darkness is 2.