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Math Help - Problem Solving

  1. #1
    Newbie
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    Sep 2008
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    Problem Solving

    Any help you give me I will appreciate greatly. Thanks!

    "
    A landscape architect needs to design a path for the Museum of Modern Art. The path is to be made of black and red tiles in the following pattern: 1 black, 1 red, 2 black, 2 red, 3 black, 3 red, 4 black, 4 red,.... The pattern of altnerating colors is to continue so that each block of a single color has one more tile than the preceding block of the same color. The museum curator estimated that 5000 tiles will be needed for the project. She ordered 2500 black tiles and 2500 red ones. If the landscape architect wants to use all of these tiles and have the path end with a block of black tiles followed by the same number of red tiles, what is the smallest number of additional black and red tiles that should be ordered?
    "

    Problems like this boggle my mind. Thanks again for any help.
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  2. #2
    Super Member

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    May 2006
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    Lexington, MA (USA)
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    Hello, nrmegin!


    It helps if you know this formula . . .

    The sum of integers from 1 to n is given by:

    . . 1 + 2 + 3 + \cdots + n \;=\;\frac{n(n+1)}{2}


    A landscape architect needs to design a path for the Museum of Modern Art.
    The path is to be made of black and red tiles in the following pattern:
    1 black, 1 red, 2 black, 2 red, 3 black, 3 red, 4 black, 4 red, ...

    The pattern of alternating colors is to continue so that each block of a single color
    has one more tile than the preceding block of the same color.

    The museum curator estimated that 5000 tiles will be needed for the project.
    She ordered 2500 black tiles and 2500 red ones.
    If the landscape architect wants to use all of these tiles and have the path end
    with a block of black tiles followed by the same number of red tiles,
    what is the smallest number of additional black and red tiles that should be ordered?

    The total number of tiles is: . T \;=\; 1 + 1 + 2 + 2 + 3 + 3 + \cdots + n + n

    . . T \;=\;2(1 + 2 + 3 + ... + n) \;=\;2\cdot\frac{n(n+1)}{2} \;=\;n(n+1)


    If T = 5000, we have: . n(n+1) \:=\:5000 \quad\Rightarrow\quad n^2 + n - 5000 \:=\:0

    Quadratic Formula: . n \;=\;\frac{-1 \pm\sqrt{20001}}{2}

    . . The positive root is: . n \:\approx\:70.212

    If n = 71, then: . T \:=\:71\cdot72 \:=\:5112


    . . Therefore, a minimum of 112 more tiles are needed.

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  3. #3
    Newbie
    Joined
    Sep 2008
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    Thanks a lot man, I really appreciate it. I wouldn't have that formula because we're doing a Problem Solving unit, that gives us no formulas. Once again, thanks a lot.
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