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Math Help - Sum of unknowns

  1. #16
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    Quote Originally Posted by Aquafina View Post
    I've edited the original question because there was a mistake in it.

    Can x and y be determined if it is given that there are 65 positive values which the sum of a non-negative multiple of x and a non-negative multiple of y cannot give

    I.e. Nx + My has 65 unattainable positive values, where N and M are non-negative integers. Also x and y are both different from 2.
    I have edited it again (in red), to indicate what I believe the problem is actually asking for. That is, you want to determine the pairs of natural numbers {x,y} such that the expression mx+ny (where m and n are non-negative integers) fails to take exactly 65 values among the natural numbers.

    Quote Originally Posted by Aquafina View Post
    I think this requires the chicken mcnugget theorem

    but i'm not sure how to use it. any ideas?

    Chicken McNugget Theorem - AoPSWiki
    There is a result stating that if x and y are coprime then the largest integer that cannot be expressed in the form mx+ny is xy–x–y. This result is apparently referred to by some people as the Chicken McNugget theorem. In more traditional circles, the number xy–x–y is known as the Frobenius number for the set {x,y}. An old result of Sylvester (quoted here) states that if x and y are coprime then the number of unattainable linear combinations of x and y is \tfrac12(x-1)(y-1).

    If we want that number to be 65 then, as you can easily see by listing all the possible factorisations of 130, the possible pairs {x,y} (with neither x nor y equal to 2) are {3,66}, {6,27} and {11,14}.
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  2. #17
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    ah i see, so if it was 64 for instance, then the answer would be

    (2,64) (4,32) and (8,16)

    and the values for x and y can be swapped around, as long as that pair is the same.. ?
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  3. #18
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    Quote Originally Posted by Aquafina View Post
    ah i see, so if it was 64 for instance, then the answer would be

    (2,64) (4,32) and (8,16)

    and the values for x and y can be swapped around, as long as that pair is the same.. ?
    Yes, except that you need to remember that these factors give you the numbers x–1 and y–1. You have to add 1 to each factor to get x and y.

    So if you are told that there are 64 unattainable numbers then the possible values of {x,y} (excluding the case when x or y is 2) are {3,65}, {5,33} and {9,17}. The braces (curly brackets) are meant to indicate that these are unordered pairs, in other words it doesn't matter which way round they are.
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  4. #19
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    thanks
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  5. #20
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    Hi how come for any value of x and y there are only a certain amount of numbers which cannot be made? Won't there ever be infinite?
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  6. #21
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    that theorem states the largest value which cannot be made is (x-1)(y-1) -1

    however, when x=2 and y =131, this would give 129... however, 265 is also unattainable and so on... all numbers greater than 131 which are not multiples of it and are odd will also be unattainable??
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  7. #22
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    Quote Originally Posted by Aquafina View Post
    Hi how come for any value of x and y there are only a certain amount of numbers which cannot be made? Won't there ever be infinite?
    As has already been said here several times, there will be infinitely many unattainable values if x and y have a common divisor greater than 1. But if x and y are coprime there are only finitely many.

    Quote Originally Posted by Aquafina View Post
    that theorem states the largest value which cannot be made is (x-1)(y-1) -1

    however, when x=2 and y =131, this would give 129... however, 265 is also unattainable and so on... all numbers greater than 131 which are not multiples of it and are odd will also be unattainable??
    265 = 67\times2 + 1\times131
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