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Math Help - Fishermen

  1. #1
    MHF Contributor harish21's Avatar
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    Fishermen

    Three fisherman come back from a day of fishing and put their fish together and go to bed. The 1st one wakes up goes to the pile of fish, throws one away and takes a 3rd and leaves. The 2nd one does the same. The last one does the same as well. What is the minimum number of fish that this can work with?
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  2. #2
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    Hello, harish21!


    Three fisherman come back from a day of fishing and put their fish together and go to bed.
    The 1st one wakes up, goes to the pile of fish, throws one away and takes a 3rd and leaves.
    The 2nd one does the same. The last one does the same as well.
    What is the minimum number of fish that this can work with?
    Let N = total number of fish.


    The 1st discards one fish and take a third of the remainder.
    Then N must be of the form: . 3A + 1
    . . N \:=\:3A + 1 .[1]

    He discards one fish . . . 3A fish left.
    He take a third of the fish ( A fish) and leaves.
    There are 2A fish left.


    The 2nd discards one fish and takes a third of the remainder.
    Then 2A must be of the form 3B + 1
    . . 2A \:=\:3B+1 .[2]

    He discards one fish . . . 3B fish left.
    He take third of the fish ( B fish) and leaves.
    There are 2B fish left.


    The 3rd discards one fish and takes a third of the remainder.
    Then 2B must be of the form 3C + 1
    . . 2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2} .[3]


    Substitute [3] into [2]: . 2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4} .[4]

    Substitute [4] into [1]: . N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}

    We have: . N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}

    Hence: . N \;=\;6C+4 + \frac{3(C+1)}{4}


    Since N is an integer, C+1 must be divisible by 4.
    The first time this happens is when C = 3.


    Therefore: . N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}

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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, harish21!


    Let N = total number of fish.


    The 1st discards one fish and take a third of the remainder.
    Then N must be of the form: . 3A + 1
    . . N \:=\:3A + 1 .[1]

    He discards one fish . . . 3A fish left.
    He take a third of the fish ( A fish) and leaves.
    There are 2A fish left.


    The 2nd discards one fish and takes a third of the remainder.
    Then 2A must be of the form 3B + 1
    . . 2A \:=\:3B+1 .[2]

    He discards one fish . . . 3B fish left.
    He take third of the fish ( B fish) and leaves.
    There are 2B fish left.


    The 3rd discards one fish and takes a third of the remainder.
    Then 2B must be of the form 3C + 1
    . . 2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2} .[3]


    Substitute [3] into [2]: . 2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4} .[4]

    Substitute [4] into [1]: . N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}

    We have: . N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}

    Hence: . N \;=\;6C+4 + \frac{3(C+1)}{4}


    Since N is an integer, C+1 must be divisible by 4.
    The first time this happens is when C = 3.


    Therefore: . N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}


    Awesome Soroban! I was given this problem in one of my classes and found it very interesting. I thought it would be nice to share. 25 is indeed the logical answer. However -2 is also a correct answer to this problem. I forgot the name of the scientist who gave an answer of -2 to this problem!!
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  4. #4
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    If the total number of fish is 25, then:

    The first person throws one away (total down to 24) and takes a third (total down to 16).

    The second person throws one away (total down to 15) and takes a third (total down to 10).

    The third person throws one away (total down to 9) and takes a third (total down to 6).

    If every time someone takes a third of the fish, the number remaining must be a whole number, then 25 is the minimum number. Without that restriction, the minimum number of fish is 4.75. In both cases I assume you're not allowed to have a negative number of fish.
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