Results 1 to 13 of 13

Math Help - Coconut Problem - Challenging

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Coconut Problem - Challenging

    The Ultimate Puzzle Site - Brain-Teasers

    Same question as there, but this problem only deals with 3 sailors and instead of splitting 1/5, they split 1/3 each time. Does this method work? If so, can someone help me complete it; I don't like their way of solving it.

    Let n = total # of coconuts. Then,

    (n - 1) - 1/3(n - 1) is after the 1st sailor;

    Then, let k = (n - 1) - 1/3(n - 1), and after the 2nd sailor will be:

    (k - 1) - 1/3(k - 1)

    Then, let m = (k - 1) - 1/3(k - 1), and after the 3rd sailor will be:

    (m - 1) - 1/3(m - 1);

    Finally, when they all wake up and give 1 away and then split the remaining, we have to do it one more time.

    Let n = (m - 1) - 1/3(m - 1), and thus the final answer is:

    (n - 1) - 1/3(n - 1)...

    Ok, now the tricky part is substituting that in. I have a whole white board filled up with substitutions, and it gets extremely messy with parentheses etc. Can anyone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I can do the problem using the Chinese Remainder Theorem but I do not know if you heard of it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by ThePerfectHacker View Post
    I can do the problem using the Chinese Remainder Theorem but I do not know if you heard of it?
    Sounds familiar from discrete- would you be able to do the substitutions instead, assuming my method works?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by Ideasman View Post
    Sounds familiar from discrete- would you be able to do the substitutions instead, assuming my method works?
    I tried doing it, and got:

    ({(n - 1) - 1/3(n - 1)} - 1) - 1/3({n - 1) - 1/3(n - 1)} - 1)) for the first substitution, then plugging this in for m and then n etc etc by the time you're finished substituting you have like a page of algebra.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ideasman View Post
    I tried doing it, and got:

    ({(n - 1) - 1/3(n - 1)} - 1) - 1/3({n - 1) - 1/3(n - 1)} - 1)) for the first substitution, then plugging this in for m and then n etc etc by the time you're finished substituting you have like a page of algebra.
    It works with the algebra approach, I know understand what you meant. Know you need to clean up the mess and then find the smallest n that makes the fraction a whole number. Which leads to solving a congruence (ax = b (mod n) ).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by ThePerfectHacker View Post
    It works with the algebra approach, I know understand what you meant. Know you need to clean up the mess and then find the smallest n that makes the fraction a whole number. Which leads to solving a congruence (ax = b (mod n) ).
    Only problem is with this approach I get lost half-way through the substitutions having like 50 different parentheses.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ideasman View Post
    Only problem is with this approach I get lost half-way through the substitutions having like 50 different parentheses.
    Parentheses never give no trouble even when there are many of them, because numbers are my friends. But there is one thing in group theory, if you ever take it, that I would imagine greatly confuses students even more than many paranthesis. It is called the factor group of a factor group. To explain basically you can think of a group as a set. A factor group is a set of sets. For example {{1,2,3},{4,2,1}}
    Now the factor group of that is the set of sets of sets.
    Thus,
    {{{1,4},{1,2,3}},{{3,5},{4,5,6}}}
    And you need to find how they add together.
    I would imagine that gives students great difficutly.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by Ideasman View Post
    Only problem is with this approach I get lost half-way through the substitutions having like 50 different parentheses.
    Aha! The power of computers. Ok, so I got:

    [[[16/81*n-16/81]-8/27]-4/9]-2/3
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Sep 2006
    Posts
    221
    Now how would I find the smallest n that makes it a whole number?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by Ideasman View Post
    Aha! The power of computers. Ok, so I got:

    [[[16/81*n-16/81]-8/27]-4/9]-2/3
    Which is equivalent to (16n)/81 - 130/81
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ideasman View Post
    Which is equivalent to (16n)/81 - 130/81
    Thus, you want to solve,
    16x\equiv 130 \mod 81
    We can divide by two because \gcd(2,81)=1,
    8x\equiv 75 \mod 81

    Using continued fractions we get that x=60 is the smallest positive integer which satisfies this congruence.

    Since this answer is not correct means that your formula that you gave me is wrong.
    Last edited by ThePerfectHacker; January 26th 2007 at 06:57 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, Ideasman!

    Ah, the "Monkey and the Coconuts" problem with three sailors.

    I have an algebraic approach to this problem . . . lengthy but simple.


    Let N = number of coconuts.

    Sailor #1 divides N coconuts by 3 and there is a remainder of 1.
    . . N \:=\:3A + 1 [1]
    He hides his A coconuts and puts 2A coconuts back in the pile.

    Sailor #2 divides 2A coconuts by 3 and there is a remainder of 1.
    . . 2A\:=\:3B + 1\quad\Rightarrow\quad A \:=\:\frac{3B + 1}{2} [2]
    He hides his B coconuts and puts 2B coconuts back in the pile.

    Sailor #3 divides 2B coconuts by 3 and there is a remainder of 1.
    . . 2B \:=\:3C + 1\quad\Rightarrow\quad B\:=\:\frac{3C+1}{2} [3]
    He hides his C coconuts and puts 2C coconuts back in the pile.

    In the morning, they divide 2C coconuts by 3 and there is a remainder of 1.
    . . 2C\:=\:3D + 1\quad\Rightarrow\quad C \:=\:\frac{3D+1}{2} [4]


    In [4]: since C is an integer, D must be odd: . D \:=\:2p-1
    Substitute: . C\:=\:\frac{3(2p-1)+1}{2}\:=\:3p - 1

    Substitute into [3]: . B \:=\:\frac{3(3p-1) + 1}{2}\:=\:\frac{9p-2}{2}
    . . Since B is an integer, p must be even: .  p\,=\,2q
    Substitute: . B \:=\:\frac{9(2q) - 2}{2}\:=\:9q - 1

    Substitute into [2]: . A\:=\:\frac{3(9q-1) + 1}{2}\:=\:\frac{27q - 2}{2}
    . . Since A is an integer, q must be even: . q \,=\,2r
    Substitute: . A\:=\:\frac{27(2r)-2}{2}\:=\:27r-1

    Substitute into [1]: . N \:=\:3(27r - 1) + 1\:=\:81r - 2


    For the least positive value of N, let  r = 1.

    Therefore: . \boxed{N \,=\,79}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    79 \div 3 \:=\:26,\text{ rem.1}

    52 \div 3 \:=\:17,\text{ rem.1}

    34 \div 3 \:=\:11,\text{ rem.1}

    22 \div 3\:=\:7,\;\text{ rem.1}

    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by ThePerfectHacker View Post
    Thus, you want to solve,
    16x\equiv 130 \mod 81
    We can divide by two because \gcd(2,81)=1,
    8x\equiv 75 \mod 81

    Using continued fractions we get that x=60 is the smallest positive integer which satisfies this congruence.

    Since this answer is not correct means that your formula that you gave me is wrong.
    If it's wrong, how is it that Soroban came up 79, which ironically makes my expression an integer? I don't think it's coincidence.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. challenging rate problem
    Posted in the Algebra Forum
    Replies: 9
    Last Post: September 4th 2011, 05:04 AM
  2. Challenging Problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 22nd 2009, 11:18 AM
  3. Challenging Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 15th 2009, 09:53 AM
  4. Challenging problem
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 13th 2009, 05:49 AM
  5. challenging sequences problem
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: February 26th 2009, 04:41 PM

Search Tags


/mathhelpforum @mathhelpforum