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Thread: Chinese remainder theorem

  1. #1
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    Montreal, Canada
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    Chinese remainder theorem

    Hello everyone.
    This is not really a homework but it is very urgent.
    I have a Discrete math exam in one day and this problem is giving me so much headache!

    $\displaystyle x\equiv 1(\bmod 4)$ (1)
    $\displaystyle x\equiv 2(\bmod 5)$ (2)
    $\displaystyle x\equiv 3(\bmod 7)$ (3)

    I tried finding inverses as well as reducing these three equations to two equations like:

    $\displaystyle x+3\equiv 0(\bmod 28)$
    $\displaystyle x+3\equiv 0(\bmod 5)$

    and in both cases the result does not work out for the equation (3).
    Could anyone please explain me why it is so and how to do this in a right way?
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    Chaneysville, PA
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    The LCD is (4)(5)(7)=140

    I like to do it like this:

    Divide by the 'mods':

    140/4=35, 140/5=28, 140/7=20

    Now, find the inverses.

    $\displaystyle 35p_{1}\equiv 1(mod \;\ 4)\Rightarrow p_{1}=3$

    $\displaystyle 28p_{2}\equiv 1(mod \;\ 5)\Rightarrow p_{2}=2$

    $\displaystyle 20p_{3}\equiv 1(mod \;\ 7)\Rightarrow p_{3}=6$

    Now, multiply:

    (1)(35)(3)+(2)(28)(2)+(3)(20)(6)=577

    $\displaystyle x\equiv 577(mod 140)$

    $\displaystyle \boxed{x=17}$

    Check the solution here:

    Javascript Calculator

    or the old-fashioned way:

    $\displaystyle \frac{17-1}{4}=4$
    $\displaystyle \frac{17-2}{5}=3$
    $\displaystyle \frac{17-3}{7}=2$

    Check.
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  3. #3
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    Thank you very much! It turns out I made a miscalculation... few times in a row . {shame on me} But anyway, thank you for your answer and for you time.
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