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A monk problem

  1. Feb 16, 2007 #1
    I saw this the other day, it's quite an interesting little problem :smile:

    If you already know it then wait 'til others have got it, in any case use spoiler tags.

    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2
    Just a quick addendum:

    The messenger, while addressing the monks at breakfast (where he can see them all), tells the monks that he can see that at least one monk already has a dot!

  4. Feb 16, 2007 #3
    Oh yeah thanks DaveE.
  5. Feb 25, 2007 #4
    None of them died, since there is no possible way of them for knowing if they have the disease or not.
  6. Feb 25, 2007 #5
    Not right, as the messenger tells them at least one monk has it, therefore one of them at least must have a dot on his head. They can all see the man in question, this means that the man who sees no red dot would be the infected one, and he'd know it.
  7. Feb 25, 2007 #6
    That much is easy, but why did he take 10 days to figure it out?
  8. Feb 25, 2007 #7
    They don't, each man dies within two hours of knowing he has the disease. After 10 days x amount of monks have died. Your making the assumption that they only die on the tenth day which is obviously not true, given the 2 hour criteria.

    Mind you I made the same assumption when I first read it.
    Last edited: Feb 25, 2007
  9. Feb 25, 2007 #8


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    Er, all the people who die do so simultaneously.
  10. Feb 25, 2007 #9
    Er no, all the people who die, die from day 1 to day 11. After they realise they are infected.

    Put it this way the priests know that someone at least is infected if two are infected, how many die? Then subsequently at each meal time how many die? Give the answer as any number you like. How many of the monks would know they had the disease?

    With 3 infected on the first morning? With 4, with 5 and so on. You have to think carefully about what the monks would think. There is only one answer.
    Last edited: Feb 25, 2007
  11. Feb 25, 2007 #10


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    This is my favorite new-knowledge problem. :tongue: If N people are infected, then all N will die shortly after the N-th time that the monks congregate.

    There's actually a very simple argument that proves all the people that die will have to die at the same time: symmetry. There is no relevant difference between any of the infected monks, so if any one if them is capable of figuring out e's infected, then all of them can.

    If two monks are infected, then nobody dies the first meal: each of the infected monks can see someone else with a dot, and thus is unsure about his own status. Then, when nobody dies after the first meal, they learn there is a second dot, and they both die after the second meal.

    If three monks are infected, then nobody dies after the second meal: at that point they know there are at least two infected people, but they all see two dots. But after the third meal...
  12. Feb 25, 2007 #11
    2 people have it, they assume that they maybe safe as only one person may have it, so one person seeing one person with the dot, may assume what if no one died? 'tis the key.:smile:
    Last edited: Feb 25, 2007
  13. Feb 25, 2007 #12


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    All of the other monks see the two people with dots die after the second meal, and conclude that they are not infected.

    Here's the timeline:

    : Messenger arrives.
    (everybody knows there is at least one dot)
    : Next meal happens.
    (everybody knows there is at least two dot)
    : The two people with dots die.
    : Next meal happens
    (everybody knows there were exactly two dots, and thus the remaining people are not infected)
    Last edited: Feb 25, 2007
  14. Feb 25, 2007 #13
    If everyone assumes there are two dots? Why? The puzzle says there are at least one person with the disease, why would they assume two and if they did what would happen? think carefully. What is the only logical solution? Considering someone at least dies on the last day in the morning?
    Last edited: Feb 25, 2007
  15. Feb 25, 2007 #14


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    (Admittedly my last post is sloppy)

    If there are two people with dots, they will die simultaneously two hours after the second meal, and nobody dies. (I count the messenger's arrival as the first meal) Are you trying to dispute this?
    Last edited: Feb 25, 2007
  16. Feb 25, 2007 #15
    Not at all, just wondering about your prognosis for the disease? What is the only logical solution?
  17. Feb 25, 2007 #16


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    Exactly what I said in post #10. And the only reason I stated the answer is because you told jimmysnyder that, in fact, the N people do not die simultaneously, which indicates you do not know the solution to the problem.
  18. Feb 25, 2007 #17
    I finally figured it out: 29 died simultaneously.
    If there were only one dot, he would have missed lunch. If there were two, they would both have missed dinner. At the end of the 10th day, all 29 dotted monks came to dinner and realized their predicament. None of them showed up for breakfast the next day.
    Last edited: Feb 25, 2007
  19. Feb 26, 2007 #18
    I found n-1 monks dying before repast n. Since breakfast on the eleventh day is the 31st repast, 30 monks will miss breakfast and be found dead in their beds. They all die simultaneously.
  20. Feb 26, 2007 #19
    No I never said x monks don't die simultaneously I simply told him that they wont all die necessarily on the 10th day, think about why?

    OK since your assuming I don't know the answer for some reason, here it is. Why would I put up a problem I didn't know the answer to? :confused:

    Hurkyl, your guilty of making assumptions that don't exist. or not accounting for something, I'm not sure which: Spoiler below.

    Here's a hint:-


    1)Don't assume the disease doesn't have an incubation period.
    2) bearing this in mind in what situation would make you know that your answer is always correct?
    3)can you assume when and on which day the disease will show up in a new person?
    4) Hurkyl does this conflict with your idea of simultaneity?


    Assuming that the only time the monks can be sure that they have it is if only one monk has it ergo he dies on the second meal. We then say one more has it and one monk is dead and so on untill the 11th day after the messenger arrived.

    However if you have 2 monks then they find out after 2 meals, if 3 after 3 meals and so on with subsequent deaths leaving potentially no monks infected, however the disease may continue to show up after the first day if it has an incubation period of 1 to x days,which of course all diseases do.

    Since we cannot be sure when the disease presents the symptoms, the only way we can be sure that at least one monk dies on the last day is if one monk dies after every meal with one infected.

    It is possible that 33 are infected but then no one else must present with the disease, 33 would then die on the last day but then the answer is the same anyway not to mention this is unlikely.

    The total is therefore the number of meals:-

    2 die on the first day
    3 for the following 10 days
    1 on the morning of the 11th day after the messenger arrives.

    Total 33.

    You have two bits of information here that you should adhere to, at least one monk has it,at least one monk dies on the last day, if 3 monks have it then it will take 3 meals before they realise that they have it, and you have to assume no one else gets it. Otherwise the system is screwed

    Essnov you're technically right like Hurkyl but you miscounted the meals.
    See above. On the morning of the 11th day after the messenger arrives.

    Spoiler part 2:-

    Your answer is technically right Hurkyl but only if 33 are infected in a precise way, otherwise you can't be sure that any more will present with the disease, so there is no way of knowing absolutely,in other words it's conditional, the other solution is much better. But well done anyway.
    Last edited: Feb 26, 2007
  21. Feb 26, 2007 #20
    i've gained 2.3 monks :rolleyes:
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