Question on "Three buildings sit next to each other"

mel106mel

The game begins:
Three buildings (A, B, and C) sit next to each other on a city block and each buidling has three floors. On each floor, the lights are either on or off.
If you recognize this game (the game date is not specified) please explain the following:
In the explanation it says that B2 is always on, whether B has all 3 or just 2 lights on. My question is, why can't B1 and B3 be on with B2 off?
If I need to write out the whole game, let me know. It's driving me crazy!!

quinnxzhang

I don't think this is an official LSAT game. Googling only turns up a test question from Kaplan.

In any case, the reason B2 is always on is because of the first three rules: (1) B has more lights on than C. (2) C1 is on. (3) Exactly two buildings have the second light on.

Because of rule (3), we're given three possibilities: A2on/B2on/C2off, A2off/B2on/C2on, and A2on/B2off/C2on. In the first two cases, B2 is on. So let's investigate the A2on/B2off/C2on case.

Suppose A2 and C2 are on and B2 is off. We know that C1 is always on from rule (2). If C2 is also on, then C has two lights on. This means B has to have all three lights on because B must have more lights on than C does, according to rule (1). But wait, we just stipulated that B2 is off. This is a contradiction. Thus, the A2on/B2off/C2on case is impossible.

Because the only two possibilities that are consistent have B2 on, B2 must always be on.

mel106mel

I knew it was going to be so obvious after it was explained. I forgot to turn on C1. Thanks for taking the time!