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. eserted island:.

  1. Aug 26, 2007 #1
    .:Deserted island:.

    A new one for you guys: a math one and a bit different version of the Gilligan's island one:

    Ten shipwrecked people land on a deserted island. There they find heaps of coconuts and a single monkey. During their first day they gather the coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning. That night one castaway wakes up hungry and decides to take his equal share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him. Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's coconut. Again, the monkey conks the man on the head and kills him. One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkey's?

    have fun!
  2. jcsd
  3. Aug 26, 2007 #2
    funny (bad) monkey---he's a NUTTER problem !!!---

    ---and that (problem) took longer than I exected
  4. Aug 26, 2007 #3
    That's a no brainer...

    The product of the greatest powers inferior to ten of the primes inferior to 10 (5*7*8*9) minus one.
  5. Aug 26, 2007 #4

    If this was a question on a test, and that was your explanation on/for your answer, you probably wouldn't score very well. (eight is a cubed number)


    and actually there is another mistake in your answer, too
    Last edited: Aug 26, 2007
  6. Aug 27, 2007 #5
    Well, I'd like a second opinion on that, it's perfectly clear and correct to my eyes.
    Last edited: Aug 27, 2007
  7. Aug 27, 2007 #6
    let's just say, the answer may be correct, but that isn't how you solve the problem
  8. Aug 27, 2007 #7
    If you say so. I'm satisfied with my answer, but I can't force you to be.
  9. Aug 27, 2007 #8
    Werg, can you explain your answer a bit more? Sadly I dont get it.
  10. Aug 27, 2007 #9
    OK, but getting an answer from the problem itself, it would be about the same as saying:

    It's from the number of letters in the words:

    In other words, there seems to be no logic or a proof from what that formula is derived or how it is formed from the problem to get the answer---if that doesn't bother you--then the answer given by the bold letters is an accepted proof too

    oh, yeah, and add one to the last word for the monkey's coconut
    Last edited: Aug 27, 2007
  11. Aug 27, 2007 #10
    Well, the number of coconuts plus one (N+1) is the minimum common multiple of 2,3,4,5,6,7,8,9 and 10.
    Then, (N+1) = (2**3) * (3**2) * 5 * 7 = 2520
    So, N=2519
  12. Aug 27, 2007 #11
    Really, rewebster, why do you want me to make this uselessly long? You know there's only one way to solve this and you know I solved it the right way. I didn't use magic to get the answer.
  13. Oct 15, 2007 #12
    The number should be something which when divided by 2..10 gives remainder 1,2,3..9, right?
    That's LCM of (2,..10) -1
  14. Oct 16, 2007 #13
    Are you saying that 8 is a prime?
  15. Oct 16, 2007 #14
    No, he said 8 is one of the greatest powers inferior to ten of the primes inferior to 10.

    And 8 is, de facto, the greatest power of 2, inferior to 10.
    It is OK.
  16. Oct 17, 2007 #15
    Just to clarify:

    1) Find all primes less than 10: 2, 3, 5, 7.
    2) Find the greatest powers of those primes such that the result is less than ten: 2^3 = 8, 3^2 = 9, 5^1 = 5, 7^1 = 7. Hence, 8,9,5,7.
    3) Find the product of 8*9*5*7. That's the LCM of all the numbers 1-10.
    4) Subtract 1 (the monkey's).

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