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Math Help - More Discrete Math Questions

  1. #1
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    More Discrete Math Questions

    1.
    I was wondering if I solved this correctly: 13x \equiv 1 (mod 28)

    gcd(28,13) = 13 * 2 + 2
    13 = 6 * 2 + 1

    so:
    1 = 13 - (6 * 2)
     1 = 13 - (6 * (28 - 13*2))
    1 = 13 - (6*28 - 13*(2*6)) = 13 - (6*28 - 13*12)
    1 = 13 - 6*28 + 13*12
    1 = 13*13 - 6*28

    13*13 \equiv 1 (mod 28)
    -------
    2.
    You have a bag of pink, yellow and green balls (woot).
    You have to pick 14 balls to be sure of at least 2 pink.
    You have to pick 12 balls to be sure of at least 2 yellow.
    You have to pick 8 balls to be sure of at least 2 green.

    How many of each?

    I thought of this as:
    yellow+green = 12
    pink+green = 10
    pink+yellow = 6

    Add them all together: 2*pink + 2*green + 2*yellow = 28. Divide by 2, and you get 14 balls total.
    If yellow+green = 12, then there are 2 pink.
    If pink+green = 10, then there are 4 yellow.
    If pink+yellow = 6, then there are 8 green.

    Did I do this correctly?
    -----
    3.
    An academic department rates its graduate program applicants numerically. Six different factors are taken into account. These different factors are weighted differently, with the integral weights chosen to add up to 100. The weight for each individual factor is at least 10. -> In how many ways could the department choose the weights?

    Each department has at least 10, so you're distributing 100-(6*10) = 40 points among six factors, however, the answer key I have says C(45,5), but I thought it should be C(45,6) or C(40+6-1,6).

    Thanks for your help
    Last edited by DemonEyesBob; December 14th 2008 at 05:05 PM.
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  2. #2
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    Lexington, MA (USA)
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    Hello, DemonEyesBob!

    13x \equiv 1 \text{ (mod 28)}
    I used a primitive approach (and got the same answer).


    We have: . 13x - 1 \:=\:28k\,\text{ for some integer }k

    Solve for x\!:\;\;x \:=\:\frac{28k + 1}{13} \:=\:2k + \frac{2k+1}{13}

    Since x is an integer, (2k+1) is divisible by 13.

    The first time this happens is: k=6

    Therefore: . x \:=\:2(6) + 1 \:=\:13 \quad\Rightarrow\quad x \equiv 13 \text{ (mod 28)}




    2. You have a bag of pink, yellow and green balls.
    You have to pick 14 balls to be sure of at least 2 pink.
    You have to pick 12 balls to be sure of at least 2 yellow.
    You have to pick 8 balls to be sure of at least 2 green.

    How many of each color?

    I thought of this as: . \begin{array}{ccc}\text{yellow + green} &=&  12 \\ \text{pink + green} &=& 10 \\ \text{pink + yellow} &=& 6 \end{array} . . . . so did I!

    I just solved the system: . \begin{array}{ccc}Y + G &=&12 \\ P + G &=& 10 \\ P + Y &=& 6 \end{array}

    . . and got: . (P,Y,G) \;=\;(2,4,8)

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  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
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    Thanks for confirming my first two answers Soroban. I don't suppose you could help with the third part? Scratch that, I figured it out. I'm set for this thread I think. I'd still appreciate help with the other thread if people can do that recurrence relation proof or the graph theory question.
    Last edited by DemonEyesBob; December 14th 2008 at 06:30 PM.
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