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Math Help - Last Project Problem Help Please!!!

  1. #1
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    Thumbs up Last Project Problem Help Please!!!



    Consider a collection of piles of bananas consisting of one pile of 9 bananas, one pile of 6, and one pile of 2. Such a collection, denoted as C1, could be expresswed as (9, 6, 2). Obtain a new collection, C2, by harvesting C1, where harvesting is defined to mean remove one banaba from each pile to form a new pile. Thus we have C2 = (8, 5, 3, 1), and if C2 is harvested we obtain:

    C3 = (7, 4, 4, 2).

    a) Let C1 = (8, 5, 2) and determine C100.

    b) Let C1 = (7, 6, 5) and determine C1995





    THANKS IN ADVANCE FOR YOUR HELP!!!!
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  2. #2
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    earboth's Avatar
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    Hi,

    please don't double-post again.

    You're wasting the time of the helpers.
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  3. #3
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    Hello, rlarach!

    I don't know of any formula to solve these,
    . . but I got them by some primitive Brute Force.


    Consider a collection of bananas: a pile of 9 bananas, a pile of 6, and a pile of 2.
    Such a collection, denoted as: C_1 \:=\:(9,\, 6,\, 2).

    Obtain a new collection, C_2, by harvesting C_1, where harvesting is defined
    . . to mean remove one banana from each pile to form a new pile.
    Thus we have: . C_2 \:= \:(8,\,5,\,3,\, 1).

    And if C_2 is harvested, we obtain: . C_3 \:= \:(7,\,4,\,4,\,2).

    a) Let C_1 \:= \:(8,\,5,\,2). .Determine C_{100.}

    b) Let C_1 \:= \:(7,\,6,\,5). .Determine C_{1995}
    (a) I listed the first few cases and found a pattern . . .

    \begin{array}{cccc}C_1 & = & (9,6,2) \\ C_2 & = & (8,5,3,1) \\ C_3 & = & (7,4,4,2) \\ C_4 & = & (6,4,3,3,1) & \leftarrow\qquad\\ C_5 & = & (5,5,3,2,2) & \uparrow \\ C_6 & = & (5,4,4,2,1,1) & \uparrow \\ C_7 & = & (6,4,3,3,1) & \text{the same as }C_4\end{array}

    After the first three steps, the sequence enters a three-step cycle.

    We have: . 100 - 3 \:=\:97\quad\Rightarrow\quad 97 \:=\:3(32) + {\color{red}1}

    Hence, C_{100} equals the first term of the cycle.

    Therefore: . C_{100} \:=\:C_4\;=\;\boxed{(6,3,3,1)}


    (b) I did the same here . . .

    \begin{array}{ccccc}C_1&=&(7,6,5) \\ C_2&=&(6,5,4,3) \\ C_3&=&(5,4,4,3,2) & \leftarrow\qquad\\ C_4 &=&(5,4,3,3,2,1) & \uparrow\\ C_5&=&(6,4,3,2,2,1) & \uparrow \\ C_6&=&(6,5,3,2,1,1) & \uparrow \\ C_7&=&(6,5,4,2,1) & \uparrow\\ C_8&=&(5,5,4,3,1)  & \uparrow\\ C_9 &=&(5,4,4,3,2) & \text{the same as }C_3 \end{array}

    After the first two steps, the sequence enters a six-step cycle.

    We have: . 1995-2\:=\:1993\quad\Rightarrow\quad 1993 \:=\:132(6) + {\color{red}1}

    Hence, C_{1995} equals the first term of the cycle.

    Therefore: . C_{1995} \:=\:C_3\:=\:(5,4,4,3,2)

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