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Math Help - fibonacci puzzle problem

  1. #1
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    fibonacci puzzle problem

    Three men possess a pile of money, their shares being , 1/3, 1/6. Each man takes some money from the pile until nothing is left. The first returns of what he took, the second 1/3, and the third 1/6. When the total so returned is divided equally among the men, it is found that each posses what he is entitled. How much money was in the original pile and how much did each man take from the pile?
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  2. #2
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    Hello, lauren2988!

    I did not come up with a unique solution . . .


    Three men possess a pile of money, their shares being \tfrac{1}{2},\:\tfrac{1}{3},\:\tfrac{1}{6}
    Each man takes some money from the pile until nothing is left.
    The first returns \tfrac{1}{2} of what he took, the second \tfrac{1}{3}, and the third \tfrac{1}{6}.
    When the total so returned is divided equally among the men,
    it is found that each posseses what he is entitled.
    How much money was in the original pile?
    How much did each man take from the pile?

    Let the three men be: . A,B,C.
    . . And they took a,b,c dollars, respectively.

    A returned half of his money.
    . . He kept \tfrac{a}{2} and returned \tfrac{b}{2} dollars.

    B returned a third of his money.
    . . He kept \tfrac{2}{3}b and returned \tfrac{b}{3} dollars.

    C returned a sixth of his money.
    . . He kept \tfrac{5}{6}c and returned \tfrac{c}{6} dollars.


    The total returned is: . \frac{a}{2} + \frac{b}{3} + \frac{c}{6} \:=\:\frac{3a+2b+c}{6} dollars.

    This was divided equally and returned to the men.
    . . They each had: . \frac{3a+2b+c}{18} dollars returned.


    A has: . \frac{a}{2} + \frac{3a+2b+c}{18} . which is equal to his original share: . \frac{a+b+c}{2}

    . . This simplifies to: . 3a - 7b - 8c \:=\:0 .[1]


    B has: . \frac{2}{3}b + \frac{3a+2b+c}{18} . which is equal to his original share: . \frac{a+bc}{3}

    . . This simplifies to: . 3a - 8b + 5c \:=\:0 .[2]


    C has: . \frac{5}{6}c + \frac{3a+2b+c}{18} .which is equal to his original share: . \frac{a+b+c}{6}

    . . This simplifies to: . b - 13c \:=\:0 .[3]


    From [3], we have: . b \,=\,13c

    Substitute into [1] and [2] and both of them become: . a \,=\,33c

    Hence, there is no unique solution.


    We have: . \begin{Bmatrix}a &=& 33c \\ b &=& 13c \\ c&=&c\end{Bmatrix}


    On the right, replace c with a parameter t.

    . . and we have: . \begin{Bmatrix}a &=& 33t \\ b &=& 13t \\ c &=& t\end{Bmatrix}


    The total amount of money is: . 33t + 13t + t \:=\:47t dollars.

    Since the original amount was divisible by 6, t must be a multiple of 6: . t \,=\,6k

    And the parametric equations become: . \begin{Bmatrix}a &=& 198k \\ b &=& 78k \\ c &=& 6k\end{Bmatrix}


    The total amount of money is: . 282k dollars ... for any positive integer k.

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