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Math Help - modular arithmetic, euclid

  1. #1
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    modular arithmetic, euclid

    using euclids algorithm compute the sum 8/122 + 11/328.
    i missed the lecture so i dnt have a clue but theres 2 methods to do this i think.i dont know what the lecturer meant by using ' / '

    does this mean 8mod122 + 11mod328
    duno how id approach this i thought its possible to say.
    8mod122=130 , 11mod328=339. 130 + 339 =469 looks very wrong.
    help?lol.
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  2. #2
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    Quote Originally Posted by sheep99 View Post
    using euclids algorithm compute the sum 8/122 + 11/328.
    i missed the lecture so i dnt have a clue but theres 2 methods to do this i think.i dont know what the lecturer meant by using ' / '

    does this mean 8mod122 + 11mod328
    duno how id approach this i thought its possible to say.
    8mod122=130 , 11mod328=339. 130 + 339 =469 looks very wrong.
    help?lol.
    Useless post, sorry.
    Last edited by Mush; December 30th 2008 at 11:04 AM. Reason: SEe above.
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  3. #3
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    does this mean 8mod122 + 11mod328
    You can't add numbers modulo different numbers.

    using euclids algorithm compute the sum 8/122 + 11/328.
    I think the notation is probably meant to mean 8 divided by 122 + 11 divided by 328. The Euclidean algorithm can be used to find the lowest common multiple of 122 and 328.
    Notations for the algorithm sometimes differ, but I do mine like this.

    328-2\times122 = 84
    122-1\times 84 = 38
    84-2\times 38 = 8
    38-4\times 8 = 6
    8-1\times 6 = 2
    6-3\times 2 = 0
    So 2 is the highest common factor.
    What is happening here is on each step we have a-n\times b = c.
    a and b are known, so we find n as the largest integer so that a-nb is positive and c as a-nb. Then we use b for the new a and c for the new b and repeat until we get 0. The highest common factor is then the value for b in the last step.
    The lowest common multiple is then \frac{122\times328}{2} , and I am sure that you know how to add fractions from this point on.
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