The other day, I saw one rather interesting movie. It was entitled

*La habitacion de Fermat*, and yes, it was in Spanish. It is a thriller (some sources say it is a horror flick, but I never saw any gory sequences) where in a small room, four mathematicians get trapped and they had to solve various mathematical puzzles within one minute, otherwise, the room gets smaller and so they would get crushed.

So, I enjoyed watching the movie, and the character development was good too. I suppose I like these psychological thrillers once in a while, since in this flick, all the action occurs within the emotional and mental dynamics of the various characters, since the film reveals various connections between the players, something that one does not expect right at the outset of the film.

So, here are some puzzles to solve.

1. If you have two hourglasses, one that is 4 minutes long and the other being 7 minutes long, how can you precisely measure a temporal length of 9 minutes?

2. Suppose there is a room that has a lightbulb inside, and three switches outside. One cannot turn the switches on or off without shutting the door. How can one determine which switch activates the lightbulb without entering the room more than once?

3. Suppose there are three boxes of sweets. One box contains mint, one box contains caramel, and the third box contains a mixture. They are all labeled, but they are all labeled incorrectly. What is the minimum number of tastes that one should do in order to determine the correct contents of the three boxes?

*DC Memorials Series*)

Ah, I love this post...

ReplyDelete-1-

I solved this in less than a minute but the only problem is that I can start measuring time only after 7 minutes have passed... is that allowed? :D. Lets call 4-minute hour glass as F and 7-minute one as S.

Start both hour glasses (t'=0). When F ends, flip it (t'=4). When S ends (t'=7) thats when you start counting. At this point, F has 1 minute left.

Measure that 1 minute, flip F, and flip again, you'll get 1+4+4 = 9

-2-

I knew this before so it took me 10 seconds to recall the solution.

-3-

Is this even a puzzle? LOL You need to taste only once. Lets call the boxes M, C, X.

Suppose you open C and it turns out to be M. Since all three boxes are labeled incorrectly, the box M must then have X and the box X must have C. There is no other possibility. QED.

My question to you: How many tastes would it take if only 1 box was correctly labeled and other 2 were not?

ReplyDeleteFinal Transit,Hmmm, you could do better with the first question. And yes, the third question is indeed correct. And I have to sit down and think about the variation on the third question.