a sinking ship puzzle I can't seem to figure out

Discussion in 'The Lounge' started by ShockTroop, Mar 9, 2007.

  1. ShockTroop

    ShockTroop Specialist

    Here's an assignment I got for a class:
    http://i7.photobucket.com/albums/y290/ShockTroop9000/IMG_00802.jpg

    As you can tell by the last paragraph, it's programming, but that's not the issue (the programming part isn't to figure it out, just simulate the process). I just can't seem to figure out what n is. I tried every X position as n and progressed clockwise, but none of them work. I also tried not skipping the ones already X'd. I haven't tried counterclockwise yet, and I've been trying to come up with a formula (we also need an explanation, "guess and test" doesn't cut it) with no luck.
    Since this is an "ancient puzzle" I tried to find some sort of mathematical explanation behind the original. I found a "puzzle of the thirty men" problem here, which gives 9 as the answer, but no explanation (the arrangement is different, too, but even if that was the answer I'd need to explain how I got it). Further searches about this just gave me the biblical story, which I would guess the problem evolved from. I also stumbled upon the Josephus problem, but that's in reverse order given that he knew the spot to stand in and n is already given. I also found a couple Japanese references but couldn't find anything.

    Now I'm not looking for an answer, just some sort of strategy on how to find a method of solving it. Would anyone happen to if the Jonah sinking ship thing the "ancient puzzle", or if it's something else? Is anyone familiar with this type of problem? It would seem easy enough, but I'm sure there's some trick to it I'm missing.

    Thanks!
     
  2. laurieB

    laurieB MajorGeek

    eliminate the nos it cannot be-:
    1,2,3,4,5,10,11,14,16,17,19,21,24,27,28. also
    6,7,8,9,12,15,23,29,
    which leaves :-
    13,18,20,22,25,26,
    the answer is thirteen. omit those x's from the next count. it had to be thirteen a because '13 is the unlucky number'
     
  3. Maxwell

    Maxwell Folgers

    Why can't it be 2? In other words every second person (i.e., those at the even numbered positions) gets thrown out thus exactly half are removed leaving 15 survivors. Doing this you don't even have to lap the circle.

    oops misread the question you have to use your layout of people.
     
  4. laurieB

    laurieB MajorGeek

    because the first five stay on board...i think you have to find a number that keeps all the 'o's and throws all the 'x's.
     
  5. laurieB

    laurieB MajorGeek

    you must have miscounted
     
  6. ShockTroop

    ShockTroop Specialist

    confused What the...I was about to say 13 was the closest, but now when I'm about to go to bed it suddenly decides to work. How did I manage to miscount several times?!?! I skipped the ones already X'd and everything...geez...
    Now a couple of my classmates are telling me the same thing. Wow...that's just weird. Holy crap.:eek:
    Thanks for, uh, confirming my consistent falseness and stupidity...or something. This is why I try hard not to make threads. :p
     

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