Every cat in Wonderland is either wise or mad. If a wise cat happens to be in the same room with three mad cats, it becomes mad. If a mad cat happens to be in the same room with three wise cats, it is exposed by them as mad. Three cats entered an empty room. Soon after the fourth cat entered, the first one went out. After the fifth cat entered, the second one went out. Etc. After the 2012th cat entered, it happened for the first time that one of the cats was exposed as mad. Which of these cats could have both been mad after entering the room? So let's case on whether the 2012th cat was wise or mad. So if the 2012th cat was mad, that means that when it entered the room, there had to be three other wise cats also in the room, which means that the cats 2011, 2010, and 2009 have to be all wise. Well, now let's see what the 2008th cat has to be. Well, if it was mad, it would have been exposed, so it had to have been wise as well, and we can just push this back to get wise cats all the way down. The thing is, each one of our answer choices is looking for two different mad cats, and we've only got one mad cat in this scenario, so that means the 2012th cat has to be wise. Okay, so let's just write that down. The 2012th cat is wise, and now that means whenever it entered the room, one of these three cats right here had to be exposed as mad. So that means exactly one of them is mad, and the other two are wise. So we have a scenario like this. Either this one is mad, and the other two are wise, or this one is mad, and the other two are wise, or this one is mad, and the other two are wise. Now let's see if we can figure out what the other previous cats were in each of these scenarios. Okay, so let's look at the 2008th cat. If it was wise, then the 2011th cat would have been exposed as mad, so that's impossible, so this guy has to be mad. Well, by exactly the same logic, these other two cats in these two scenarios have to be mad as well. Now let's look at the 2007th cat. In this scenario right here, we're looking at WWM, which, conveniently enough, we've seen right here with WWM, and so we know how to push that back by one cat, so let's just make it mad. Now in this scenario, we have mad, wise, and mad. If this cat was mad, then this cat would have become mad, and that's a problem because we know that it stayed wise throughout. Okay, so that means that this cat has to be wise, and by the same reasoning, this cat has to be wise in this last scenario as well. Well, now we see a WMM, and conveniently enough, we've already seen that down here, so we know how to push that back by one more cat to get a W here, and then we can use similar reasoning to deduce these other two rows of cats as being wise, mad, wise, and then wise, wise, mad, but now take a look at this grid right here. This is exactly the same as what we started with, and so that means every time we move back by four cats, we end up in exactly the same scenario as where we were before, so that means the only columns in this table that really matter are these four columns, and we can figure out all the other columns by just going back in multiples of four, so let's just throw out these because we only care about these. Now we care about columns one, two, three, and four because those are the ones that show up in the answer choices. Well, let's just see which one of them they match up to. Well, this one has to match up to four because this is a multiple of four. This one has to match up to one, and then two, and then three. All right, now let's just see which cats could have both been mad at the same time, so can the first cat and the 2011th cat be mad at the same time? So the 2011th cat, we need to look in this row for the first cat in this column. For the first cat, we need to look in this column. In this scenario, they're not both mad. In this scenario, they're not both mad, and in this scenario, they're not both mad either, so the answer isn't A. Let's try out B. Second and 2010, well, those are just the same row, and right here, we see that both cats could possibly be mad, so the answer should be B. Let's just check the other ones to make sure that they don't end up happening. If we look at the third column here and the 2009th column here, we see that the two cats aren't both mad at the same time. If we look at the fourth one and the last one, well, here's the fourth one. The last cat is just always wise, so this guy is gone, and the last choice is the second option in the 2011th, the second column in the 2011th column, and so we just have to look here and here, and both of these cats aren't both mad at the same time, so that means our answer has to be B.

Wonderland

wise cat

mad cat

2012th cat

scenario analysis