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Math Help - Modular Arithmetic

  1. #1
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    Modular Arithmetic



    Here is the solution they've given us for (a):

    We have to solve:
    7 2 + 3 1 + 9 1 + 7 8 + 3 7 + 9 2 + 7 9 + 3 4 + 9 x = 0 mod 10

    196 + 9x = 0 mod 10

    6 − x = 0 mod 10

    x = 6
    So, how did they get from "196+9x" to "6-x"? I know that 196 mod 10 = 6, but why did they change "9x" to "-x"? What did they do to it?

    Also this is the given solution to part (b):

    228 + 3x = 0 mod 10
    8 + 3x = 0 mod 10
    3x = -8 = 2 = 12 mod 10
    x = 4


    I know that 228 mod 10=8, but this time they didn't change the other term (3x)...

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  2. #2
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    9x = -x mod 10

    The terms are altered so that the coefficient of x divides the other number.
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  3. #3
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    Quote Originally Posted by demode View Post


    Here is the solution they've given us for (a):



    So, how did they get from "196+9x" to "6-x"? I know that 196 mod 10 = 6, but why did they change "9x" to "-x"? What did they do to it?
    Same thing, really, 9= 10- 1 so 9 mod 10= -1.

    Also this is the given solution to part (b):



    I know that 228 mod 10=8, but this time they didn't change the other term (3x)...

    They didn't need to. 12 is evenly divisible by 3. Had it reduced to, say 3x= 7, then they would have set 3 mod 10= -7 and had -7x= 7 so x= -1= 9 (mod 10) or 3x= -3 (mod 10).
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  4. #4
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    So in these type of problems we should always look for a way to make the coefficient of x divide the other number, or make the other number divisible by the coefficient of x?
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