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Math Help - Congruence system

  1. #1
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    Congruence system

    Hi, i need someone to explain me how to solve the following exercises, please try to be clear because i don't understand too much of it. thanks.


    the first problem is how to solve this two exercises:

    Congruence system-codecogseqn.gif


    Congruence system-codecogseqn2.gif



    and the the last one is a system of congruences

    Congruence system-codecogseqn3.gif
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  2. #2
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    Re: Congruence system

    Hey mcolula.

    What is covered in your notes?

    If you want to get some notes and algorithms, the subjects you need to look at include bi-linear diophantine equations, and the Chinese remainder theorem.
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  3. #3
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    Re: Congruence system

    Hello, mcolula!

    If you truly "don't understand too much of it",
    . . my explanations won't help you at all.


    3x+2 \:\equiv\:4\text{ (mod 7)}

    \begin{array}{cccccccc}\text{We have:} & 3x &\equiv& 2 & \text{(mod 7)} \\ \text{Multiply by 5:} & 15x & \equiv & 10 & \text{(mod 7)} \\ & x &\equiv & 3 & \text{(mod 7)} \end{array}




    2\:\equiv\:5x\text{ (mod 3)}

    \begin{array}{ccccccc}\text{We have:} & 5x & \equiv & 2 & \text{(mod 3)} \\ & 2x & \equiv & 2 & \text{(mod 3)} \\ & x &\equiv& 1 & \text{(mod 3)} \end{array}




    x\:\equiv\:9\text{ (mod 7)}
    x \:\equiv\:6\text{ (mod 2)}
    x \:\equiv\:5\text{ (mod 5)}

    \begin{array}{cccccccccccc}x & \equiv & 9 & \text{(mod 7)} & \Rightarrow & x &\equiv & 2 \text{ (mod 7)} & \Rightarrow & x \:=\:2 + 7a \\ x & \equiv & 6 &\text{(mod 2)} & \Rightarrow & x & \equiv & 0\text{ (mod 2)} & \Rightarrow & x\text{ is a multiple of 2.} \\ x & \equiv & 5 & \text{(mod 5)} & \Rightarrow & x &\equiv& 0\text{ (mod 5)} & \Rightarrow & x\text{ is a multiple of 5.} \end{array}

    We have: . \begin{Bmatrix}x &=& 9+7a & {\color{blue}[1]} \\ x &=& 10b & {\color{blue}[2]}\end{Bmatrix}

    Then: . 9 + 7a \:=\:10b \quad\Rightarrow\quad a \:=\:\frac{10b-9}{7} \:=\:b-2 + \frac{3b-2}{7}

    Since a is an integer, 3b-2 must be a multiple of 7.
    . . 3b-2 \:=\:7c \quad\Rightarrow\quad b \:=\:\frac{7c+2}{3} \:=\:2c + \frac{c+2}{3} .[3]

    Since b is an integer, c+2 must be a multiple of 3.
    . . This happens when c \,=\,3k+1

    Substitute into [3]: . b \:=\:\frac{7(3k+1)+2}{3} \quad\Rightarrow\quad b \:=\:7k+3

    Substitute into [2]: . \boxed{x \:=\:10(7k+3)}
    Thanks from topsquark
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