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Math Help - Easy Unattainable Score Problem

  1. #1
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    Easy Unattainable Score Problem

    Hello,

    My book says that there is a simple proof to provide the following, but it doesn't give it. Can someone please explain this:

    Suppose there is a game in which there are two kinds of scoring events. One event gives a score of m points, and the other gives a score of n points. Assuming that m and n are relatively prime, we can derive a formula for the largest unattainable score and prove that the answer is correct.

    Can anyone help derive this formula and prove that its true?
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  2. #2
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    Quote Originally Posted by 1337h4x View Post
    Hello,

    My book says that there is a simple proof to provide the following, but it doesn't give it. Can someone please explain this:

    Suppose there is a game in which there are two kinds of scoring events. One event gives a score of m points, and the other gives a score of n points. Assuming that m and n are relatively prime, we can derive a formula for the largest unattainable score and prove that the answer is correct.

    Can anyone help derive this formula and prove that its true?
    I haven't looked over this too closely, but you can try this link. (Yahoo! Answers - Sylvester's Proof that g(a,b) = ab - (a+b)?)
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  3. #3
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    Quote Originally Posted by undefined View Post
    I haven't looked over this too closely, but you can try this link. (Yahoo! Answers - Sylvester's Proof that g(a,b) = ab - (a+b)?)

    I'm confused by this guy's proof. I have only m and n in my problem, he has m and n, a and b, etc.

    Could someone clarify this and make the connection to what I was asking?
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  4. #4
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    Anyone have any clarity to offer?
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  5. #5
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    Where did you see that the other poster's proof had an n? There are a, b, x, y, m, but no n.

    It is merely a choice of letter assignments, which are really arbitrary. Here your m and n are instead called a and b. When we talk of a linear combination of a and b in nonnegative integers, that can be expressed as ax + by, for nonnegative integers x and y. (Any attainable score is a linear combination of a and b.)

    m is introduced by the definition of congruence. Suppose

    c \equiv d\ (\text{mod }e)

    This is true if and only if c - d is a multiple of e; that is, there exists an integer f such that

    c - d = fe

    which is the same as, there exists an integer f such that

    c = d + fe

    I still have not looked over the lattice points part, but I did not find any errors in the proof provided by the poster regarding the first part.
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  6. #6
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    So we need to prove two things. Given positive integers a, b with gcd(a,b)=1:

    1. For all integers c > ab-a-b, there exist nonnegative integers x, y such that ax + by = c.

    2. For c = ab-a-b, there are no solutions to ax + by = c in nonnegative integers.

    - Hollywood
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