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Math Help - five men, a monkey, and a pile of coconuts

  1. #1
    Senior Member Mukilab's Avatar
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    five men, a monkey, and a pile of coconuts

    On a desert island, five men and a monkey gather coconuts all day, then sleep.
    The first man awakens and decides to take his share. He divides the coconuts into five equal shares with one left over. He gives the extra one the, hides his share and goes back to sleep. Later the second man awakens and takes his fifth from the remaining pike; he too finds one extra one and gives it to the monkey. Each of the remaining three men does likewise in turn. Find the minimum number of coconuts originally present.

    I need a full explanation + answer please.

    I tried starting off with \frac{x}{5} \cdot \frac{5}{1}

    Would I do this but adding 1 each time?
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  2. #2
    Senior Member Mukilab's Avatar
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    would this work?

    (\frac{5}{1}(\frac{5}{1}(\frac{5}{1}(\frac{5}{1}(\  frac{x-1}{5})-1)-1)-1)-1)


    dividing it by 1/5 each time and minusing 1?
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    On a desert island, five men and a monkey gather coconuts all day, then sleep.
    The first man awakens and decides to take his share. He divides the coconuts into five equal shares with one left over. He gives the extra one the, hides his share and goes back to sleep. Later the second man awakens and takes his fifth from the remaining pike; he too finds one extra one and gives it to the monkey. Each of the remaining three men does likewise in turn. Find the minimum number of coconuts originally present.

    I need a full explanation + answer please.

    I tried starting off with \frac{x}{5} \cdot \frac{5}{1}

    Would I do this but adding 1 each time?
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  4. #4
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by mr fantastic View Post

    I had the wrong numerator :/

    thanks
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  5. #5
    Senior Member Mukilab's Avatar
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    er, at first I thought this was right

    It's a different question

    I tried using the same technique on my question and it didn't work

    I need an explanation please
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  6. #6
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    Using the fact that the difference between the amount of coconuts after each step must be an integer, and the difference is multiplied by 4/5 after each step, the first difference must be divisible by 5^4 so that the last difference is an integer. The smallest such difference is 625, which gives you an initial coconut count of 3121.
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  7. #7
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by icemanfan View Post
    Using the fact that the difference between the amount of coconuts after each step must be an integer, and the difference is multiplied by 4/5 after each step, the first difference must be divisible by 5^4 so that the last difference is an integer. The smallest such difference is 625, which gives you an initial coconut count of 3121.
    that is what I got but it doesn't work if you do the calculation.

    Also, your notes are very hard to follow
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  8. #8
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    five men, a monkey, and a pile of coconuts
    wait ...

    I thought it was Churchill, Stalin, Roosevelt, and a monkey in a bar?
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  9. #9
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    Deleted- misread the problem.
    Last edited by HallsofIvy; March 13th 2010 at 01:01 PM.
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  10. #10
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    Hello, Mukilab!

    This is a classic problem: "The Monkey and the Coconuts".
    It has a long (but very elementary) solution.


    On a desert island, five men gather coconuts all day, then sleep.

    The first man awakens and decides to take his share.
    He divides the coconuts into five equal shares with one left over.
    He gives the extra one to the monkey, hides his share,
    puts the remaining coconuts in a pile, and goes back to sleep.

    Later the second man awakens and takes his fifth from the remaining pile.
    He too finds one extra one and gives it to the monkey.

    Each of the remaining three men does likewise in turn.

    Find the minimum number of coconuts originally present.
    The original problem concluded with:
    . . In the morning, the five men innocently divided the remaining coconuts.
    . . This time the division came out even; no coconut for the monkey,.


    Let N = the original number of coconuts.

    The first man divided them into 5 equal piles with one left over: . N \:=\:5A + 1 \;\;[1]
    He hid his A coconuts and put the 4A coconuts in a pile.

    The second man divided them into five piles with one left over: . 4A \:=\:5B + 1\;\;[2]
    He hid his B coconuts and put the 4B coconuts in a pile.

    The third man divided them into five piles with one left over: . 4B \:=\:5C + 1\;\;[3]
    He hid his C coconuts and put the 4C coconuts in a pile.

    The fourth man divided them into five piles with one left over: . 4C \:=\:5D + 1\;\;[4]
    He hid his D coconuts and put the 4D coconuts in a pile.

    The fifth man divided them into five piles with one left over: . 4D \:=\:5E + 1\;\;[5]
    He hid his E coconuts and put the 4E coconuts in a pile.

    In the morning, they divided them into five equal piles: . 4E \:=\:5F\;\;[6]



    From [6]: . E \:=\:\frac{5F}{4}

    Since E is an integer, F is a multiple of 4: . F \:=\:4p

    Then we have: . E \:=\:\frac{5(4p)}{4} \:=\:5p



    Substitute into [5]: . 4D \:=\:5(5p)+1 \:=\:25p + 1

    We have: . D \:=\:\frac{25p+1}{4} \:=\:6p + \frac{p+1}{4}

    Since D is an integer, p+1 must be a multiple of 4: . p+1\:=\:4q \quad\Rightarrow\quad p \:=\:4q-1

    Then we have: . D \:=\:6(4q-1) + \frac{(4q-1)+1}{4} \quad\Rightarrow\quad D\;=\;25q-6



    Substitute into [4]: . 4C \:=\:5(25q-6) + 1 \:=\:125q - 29

    Then: . C \:=\:\frac{125q-29}{4} \:=\:31q - 7 + \frac{q-1}{4}

    Since C is an integer, q-1 must be a multiple of 4: . q-1 \:=\:4r \quad\Rightarrow\quad q \:=\:4r+1

    Then we have: . C\:=\:31(4r+1)-6 + \frac{(4r+1)-1}{4} \quad\Rightarrow\quad C \:=\:125r + 24



    Substitute into [3]: . 4B \:=\:5(125r + 24) + 1 \:=\:625r + 121

    Then: . B \:=\:\frac{625r + 121}{4} \:=\:156r + 30 + \frac{r+1}{4}

    Since B is an integer, r+1 must be a multiple of 4: . r+1\:=\:4s \quad\Rightarrow\quad r \:=\:4s-1

    Then we have: . B \:=\:156(4s-1) + 30 + \frac{(4s-1)+1}{4} \quad\Rightarrow\quad B \:=\:625s - 126



    Substitute into [2]: . 4A \:=\:5(625s - 126) + 1 \:=\:3125s - 629

    Then: . A \:=\:\frac{2125s-629}{4} \:=\:781s - 157 + \frac{s-1}{4}

    Since A is an integer, s-1 is a multiple of 4: . s-1 \:=\:4t \quad\Rightarrow\quad s \:=\:4t+1

    Then we have: . A \:=\:781(4t+1) - 157 + \frac{(4t+1)-1}{4} \quad\Rightarrow\quad A \:=\:3125 + 624



    Substitute into [1]: . N \:=\:5(2125t+ 624) + 1

    . . . . and we have: . N \:=\:15625t + 3121


    The least N occurs when t = 0

    . . Therefore: . N \:=\:3121

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