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Thread: 4 skaters and a monkey

  1. #1
    Custom Title Mathman's Avatar
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    4 skaters and a monkey

    Kimmie, Mao, Yu-na and Emily were marooned on a desert island, along with Kimmie's pet monkey. With only a pile of coconuts to sustain them until they were rescued, naturally tensions rose.

    In the middle of the night, Kimmie, suspicious of her friends, got up and snuck down to the beach where the coconuts were. She wanted to make sure that she got her fair share, so she divided the coconuts into four equal piles. There was one coconut left over, which she threw to the monkey.

    Then she took her share back to her tent and hid them, leaving three quarters (less one) of the original pile on the beach.

    Then Mao woke up and did the same thing. Not knowing that Kimmie had been there before her, she snuck down to the beach and divided the remaining coconuts into four equal piles. There was one left over, which she threw to the monkey. Then she took her share back to her tent.

    Yu-na and Emily followed suit. Each divided the remaining pile by four and took away one-fourth, throwing the extra coconut to the monkey.

    How many coconuts were there to begin with?

  2. #2
    Forum translator Ptichka's Avatar
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    I must really be missing something, because I still get infinite answers:
    341, 597, 853, 1109, 1365, etc.

  3. #3
    Custom Title Mathman's Avatar
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    Maybe I didn't explain it right. After each division there must be a remainder of exactly one for the monkey.

    Try 341.

    Kimmie divides 341 by four, to get a quotient of 85 and a remainder of 1. So far so good.

    Now there are 341 - 85 -1 = 255 left.

    Mao divides 255 by 4 to get a quotient of 63 and a remainder of 3. Oops.

    You got the main point right, though. Yes, there are infinitely many solutions. Once you get the smallest one, n, the rest are n+256k, k = 1, 2, 3,...

    Do you want a hint for how to do it in a slick way without doing any calculations at all? Look for a solution where n is a single digit negative number, LOL.

    Antother slick method: If you think about it right, the four steps specify the 4 digits of n written in base 4.

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    On Edge Piel's Avatar
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    What a let down! I thought this thread was going to be about the lives and loves of Sasha Zhulin .

  5. #5
    Custom Title Mathman's Avatar
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    :chorus: :chorus:

  6. #6
    and... World Peace! Tonichelle's Avatar
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    Why do I have a feeling Piel was class clown in math classes?

    Math - bleh - who needs it?

  7. #7
    Custom Title Mathman's Avatar
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    Quote Originally Posted by Tonichelle
    Math - bleh - who needs it?
    OK, just for that I will tell you the answer.

    Solution: There were negative three coconuts on the beach.

    Staring with -3 coconuts, Kimmie gives +1 coconut to the monkey. Now there are -4. Kimmie divides the -4 coconuts by 4, so each skater's share is -1 coconut.

    Kimmie takes her -1 coconut back to her tent, leaving -3 in the pile.

    Now Mao comes along. There are -3 coconuts in the pile, just like before. She gives +1 to the monkey, divides the resulting -4 by 4, takes her -1 coconut back to her tent and leaves -3 still in the pile.

    This continues for however many skaters you have. Each skater takes away -1 coconut, the monkey gets +1, and there are always -3 left on the beach.

    And you say we don't need math!

  8. #8
    Tripping on the Podium
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    That's stupid. You can't have negative three coconuts on the beach.

    How long do you think you can live off negative three coconuts? Negative three weeks? So they all died three weeks before they got there?

  9. #9
    Custom Title Mathman's Avatar
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    This problem is really about the question, what is the remainder when you divide by 4, then divide by 4 again, etc., four times. This is the same a dividing by 4x4x4x4 = 256. So once you have one answer, you can get another by adding on any multiple of 256. So particular,

    -3 + 256 = 253 coconuts in the pile is the smallest solution in positive integers.

    Check: Start with 253 coconuts.

    Kimmie + Mao + Yu-na + Emily + monkey + still left = total.

    63 + 47 + 35 + 26 + 4 +78 = 253. (check!)

    (Don't bother to thank me, RBP -- it is my duty to post these math things once a year.)

  10. #10
    Tripping on the Podium
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    So we're safe for another year now?

  11. #11
    On Edge Piel's Avatar
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    Thank you MM for caring about our education. RBP thanks you too, in his own special way. RBP we are past due for a poem from you.

  12. #12
    and... World Peace! Tonichelle's Avatar
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    Quote Originally Posted by Mathman
    OK, just for that I will tell you the answer.

    I'm so good at getting people to do the work for me

  13. #13
    Forum translator Ptichka's Avatar
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    MM, I agree that this is not a clean problem. If you just make into a boring numbers problem, it works. However, when you are talking about people physically taking their share into their tent, it cannot be a negative number.

  14. #14
    MY TVC 1 5 SeaniBu's Avatar
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    Quote Originally Posted by Mathman
    Solution: There were negative three coconuts on the beach.
    Did everyone eat except Mao and the monkey when they got there???? the "reality" hypotenuse doesn't support the equation here does it?
    Now I know that acording to math, motion never reaches it final destination....
    1/2 way, 3/4 way, ect on to infintiy...or does that say Math doesn't suport reality???? Need a fiendish emoticon!!!! mooohaaahaaa
    Last edited by SeaniBu; 06-12-2006 at 05:11 PM.

  15. #15
    Custom Title Mathman's Avatar
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    Quote Originally Posted by Ptichka
    MM, I agree that this is not a clean problem. If you just make into a boring numbers problem, it works. However, when you are talking about people physically taking their share into their tent, it cannot be a negative number.
    OK, I will try again. Let N be the number of coconuts (a positive number), and write N in base four:

    N = a3x4^3 + a2x4^2 + a1x4^1 + a0x4^0,

    with each ai between 0 and 3.

    The first condition (Kimmie has one coconut left for the monkey) means that a0 = 1.

    The second condition (and all the rest) mean that 3/4 of what's left also has a remainder of one when divided by four, so each of the other ai's must = 3.

    So we have N = 3331 in base four, which is equal to 253 in base ten.

    All other solutions in postive integers are of the form 253 + a multiple of 256.

    So this really is a well-defined problem (if you pose it like this: what's is the smallest positive number of coconuts that can be in the pile?)

    But what is interesting (to me, LOL) is that by thinking about abstractions like negative numbers we get the right answer, 253, instantly, without having to figure out anything. Thus illustrating the power of mathematical abstraction (= making it into "just a boring number problem," LOL.)

    MM
    Last edited by Mathman; 06-12-2006 at 05:51 PM.

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