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Thread: solve the system

  1. #1
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    Post solve the system

    x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
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  2. #2
    Moo
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    hello,
    Quote Originally Posted by Sally_Math View Post
    x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
    $\displaystyle 2x \equiv 3 \bmod 7 \equiv 10 \bmod 7 \Rightarrow x \equiv 5 \bmod 7$

    $\displaystyle 3x \equiv 4 \bmod 11 \equiv 15 \bmod 11 \Rightarrow x \equiv 5 \bmod 11$

    now use the chinese remainder theorem.
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  3. #3
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    Quote Originally Posted by Moo View Post
    hello,

    $\displaystyle 2x \equiv 3 \bmod 7 \equiv 10 \bmod 7 \Rightarrow x \equiv 5 \bmod 7$

    $\displaystyle 3x \equiv 4 \bmod 11 \equiv 15 \bmod 11 \Rightarrow x \equiv 5 \bmod 11$

    now use the chinese remainder theorem.
    ok i know that u start this way:-
    using a=b+km (for eq. 1)
    which is x=2+5k
    then substituting into eq. 2, u get
    2(2+5k)=3 (mod 7)
    then we get
    4+20k=3 (mod 7)

    and then that's where i get stuck
    so please help me understand the way we finish it.
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  4. #4
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    Quote Originally Posted by Sally_Math View Post
    x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
    The congruence, $\displaystyle 2x\equiv 3(\bmod 7)$ is equivalent to $\displaystyle 4(2x)\equiv 4\cdot 3 (\bmod 7)$ which becomes $\displaystyle x\equiv 5(\bmod 7)$. The congruence, $\displaystyle 3x\equiv 4(\bmod 11)$ is equivalent to $\displaystyle 4(3x)\equiv 4\cdot 4(\bmod 11)$ which becomes $\displaystyle x\equiv 5(\bmod 11)$.

    Thus, you have to solve,
    $\displaystyle x\equiv 2(\bmod 5)$
    $\displaystyle x\equiv 5(\bmod 7)$
    $\displaystyle x\equiv 5(\bmod 11)$

    Combine the last two congruences and so,
    $\displaystyle x\equiv 2(\bmod 5)$
    $\displaystyle x\equiv 5(\bmod 77)$

    This is equivalent to,
    $\displaystyle x\equiv 2 + 16\cdot 5(\bmod 5)$
    $\displaystyle x\equiv 5 + 77(\bmod 77)$

    We get,
    $\displaystyle x\equiv 82(\bmod 5)$
    $\displaystyle x\equiv 82(\bmod 77)$

    Combine,
    $\displaystyle x\equiv 82(\bmod 385)$
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