Puzzles in Fermat’s Room

I watched Fermat’s Room — a mathematical thriller in Spanish language — few weeks ago. Although I did enjoy watching this movie, I am hesitant to recommend it due to sloppy direction (too pretentious, IMO) and average acting performances. But the premise on which the movie is based was quite intriguing: Four mathematicians are invited by an anonymous stranger (Fermat) to solve a great mathematical enigma. But soon after they convene, they get trapped in the room and must solve puzzles in order to survive. The room, with moving walls, begins to shrink as soon as they receive the puzzle from the host and it won’t stop contracting until they solve the puzzle.

Disappointingly, the puzzles were quite basic – most of them are classic riddles that I was already familiar with. I’ve listed some of them below:

  • Using a 4 minute hourglass and a 7 minute hourglass how would you measure exactly 9 minutes?
  • You have three opaque boxes. One box contains chocolate candies, another contains mint candies, and the last box contains a mixture of chocolate and mint. The boxes are labeled Chocolate, Mint and Mixed. None of the boxes are labeled correctly. You can take one candy out of each box (without looking directly into the box) and see what you get. What is the minimum number of boxes you have to open (and take one candy out) to assign correct labels to all boxes?
  • There are three switches outside a room. One of them controls a lightbulb that’s inside the room, while the other two are not connected to anything. You can turn the switches ‘on’ or ‘off’ and also enter the room to see the lightbulb. What is the fewest number of times you will need to enter the room to determine which switch is connected to the bulb? (You can’t see if the lightbulb is ‘on’ or ‘off’ from the outside. And all switches are in the ‘off’ position when you start.)
  • A mother is 21 years older than her son. In exactly 6 years, the son will be one-fifth his mother’s age. The question is: what is the father doing right now?

I think the following is probably the best one in the lot:

  • A student asks his professor: “What are the ages of your three children?” The professor replies: “If you multiply their ages you get 36, and if you add them you get my house number.” “I know your house number, but that’s not enough information!” says the student. To that the professor answered: “True. The oldest lives upstairs.” What are the ages of the three children?

And finally, a classic Martin Gardner riddle: What is special about the number 8,549,176,320?

[Picture Courtesy: IFC Films]

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18 responses to “Puzzles in Fermat’s Room

  1. Pingback: Exam [Movie] | A Blank Slate

  2. Pranav Mathur

    i knew the second and fifth ones already.. as for the fourth one, i think the father should not be disturbed right now, he’s way too busy engaging in coitus ;)!! need the answers for the first and third questions!!

    • Hi Pranav,

      Solution for the Hourglass puzzle: Let’s say A=4 minute hourglass, and B=7 minute hourglass. Follow these steps to count 9 minutes:

      (1) Start with both A and B until A is completely drained. That’s the 4 minute mark.

      At the end of 4 minutes, B has 3 minutes worth of sand left on its top half (and the bottom half has 4 minutes worth of sand). And A is completely drained – i.e. all 4 minutes worth of sand is on its bottom half.

      (2) Flip A and B, and let B (which is equal to 3 minutes) completely drain. That’s the 4+3=7 minutes mark.

      At the end of 7 minutes, now A has 1 minute worth of sand left on its top half (and 3 minutes at the bottom half). While B is completely drained.

      (3) Flip both of them again B, and let A (which is equal to 1 minute) completely drain. That’s the 7+1=8 minutes mark.

      At the end of 8 minutes, A has all its sand on one side (bottom), while B has one minute worth of sand collected at the bottom (and 7 minutes left at the top half).

      (4) Flip B, and let it drain. That’s the 8+1=9 minutes mark.

      Done!

      (Let me know if that’s unclear.)

      Solution for the Light-switch puzzle: Turn on any one switch, let’s say A. Wait for 10 minutes. Turn it off and then turn on any other key, let’s say B. Leave B on and enter the room to inspect the light-bulb.

      If one the light-bulb is on – than switch B is connected to that bulb. (Because we have left that switch on.)

      Otherwise, touch the light-bulbs and see if any of them it is warm (or still hot). If you find a light-bulb that’s it warm, then switch A is connected to that bulb. (Because that switch was left on for 10 minutes, which should have made the light-bulb warm.)

      If none of the light-bulbs are the bulb is not warm, then we must conclude that switch C is connected to a that light-bulb. (Because at least one switch is connected.)

      Note that the objective is not to identify the correct pairs of switch and light-bulbs, just to identify which switch is connected.

    • Lonnie H.

      if the father is indeed engaged in coitus. that would mean that in exactly 6 years, the child would be 5.25 years old. if the mother will be 27 in 6 years then 1/5 of that is 5.4, not 5.25 years. if it means that the child will be 1/5 of the mothers starting age of 21, then 1/5 of that would be 4.2. am i missing something?

      • Anonymous

        The mother is not 21, she is 20.25 years old (because her son, currently being conceived, is -0.75 years old). So in 6 years time she will be 26.25 years old, which is five times the age her son will be then, ie 5.25.

  3. Ed

    First of all, I already love this blog having only started reading it yesterday. But I have a couple of problems with your solutions.

    Firstly, in step 2 you should only be flipping A; B has 3 minutes left in its top half, so flipping it would measure 4 minutes.The same mistake is in step 3; you only need to flip B, because A has one minute left in the top half not the bottom half.

    Secondly, the third question makes it sound as if there’s only one lightbulb in the room, but your solution does in fact still work in this case. Sorrry to start on a critical note.

    I was wondering if you have the solution to the second problem; I couldn’t work it out. Thanks and keep up the fascinating work

    • Hi Ed,

      Glad you liked my blog! And you are correct about both of my mistakes. I will edit my earlier response (comment) accordingly.

      Here’s the solution of the second problem — and I hope you won’t find another mistake in this one… 🙂

      Correct Answer: You need to open only one box.

      Explanation: The key point is that ALL of the three boxes are labeled incorrectly.

      Start with the box labeled Mixed, and take one candy out. There are two possibilities:

      (A) You pulled out a Chocolate candy, or (B) you pulled out a Mint candy.

      If (A) is true then the correct label for that box must be Chocolate. Note that Mint can’t be the correct label for this box (obviously) because you now know that it contains at least one Chocolate candy, and it can’t be Mixed because otherwise the initial label would have been correct (which violates the “all labels are incorrect” condition.)

      And similarly, if (B) is true, then the correct label for this box must Mint.

      Once you determine the correct label for the box that you opened, the rest is easy. At that point, you have:
      X = a correctly labeled box (the one that you opened, and correctly inferred its content),
      Y = a box without any labels (this is the box from which you peeled off the label to stick it on X), and
      Z = a box with an incorrect label.

      The (incorrect) label on Z has only one place to go: Y. And the label that you peeled off from X goes to Z.

      Makes sense? 🙂

  4. Ed

    Brilliant, I completely overlooked that piece of information. Thanks a lot

  5. nick

    for the last one
    the number 8,549,176,320 it
    contains all numbers from 0 – 9 and
    it is divisable by all the numbers from 0-9

    • Nick,

      That number is actually NOT divisible by 7. The correct answer is: if you arrange all numbers from 0 to 9 alphabetically, you get this number: eight, five, four, nine, one, seven, six, three, two and zero.

  6. Anonymous

    lovely blog, thanks for all the posts!

  7. Anonymous

    Wow! i’m having a hard time with the student-professor quiz.. :(( need an answer for that fifth question, pls. 🙂 And oh! i really love your blog! :))))

    • Start by writing down all three-way factorization of 36 (because we know that x * y * z = 36).

      Now, 36 = 1 x 2 x 2 x 3 x 3, which gives us several possibilities for the ages of three children: {1,4,9}, {1,2,18}, {1,3,12}, {2,3,6}, {2,2,9}, {3,3,4}, {1,6,6} and {1,1,36}.

      Now, the sum of those three ages are: 14, 21, 16, 11, 13, 10, 13, and 38, respectively. Notice that there are two possibilities, namely {2,2,9} and {1,6,6}, for which the ages sum to 13.

      Because prof’s first two hints were *not* enough to deduce the answer, the correct answer must be one of these two cases. (Because in all other cases, the student would have been able to deduce the answer based on the first two hints.) The student knows that the prof’s house number is 13 and the two possible answers are {2,2,9} and {1,6,6} but he has insufficient information to determine which one of these two answers is correct.

      Now, the professor either has one older child and two young ones (possibly twins), or he has one younger child and two older ones (possibly twins). Prof’s third comment hints that he has one older child (who happens to live upstairs), and that helps the student solve the problem. The correct answer is: {2,2,9}.

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