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Math Help - 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)
    2x \equiv 3 \bmod 7 \equiv 10 \bmod 7 \Rightarrow x \equiv 5 \bmod 7

    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,

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

    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, 2x\equiv 3(\bmod 7) is equivalent to 4(2x)\equiv 4\cdot 3 (\bmod 7) which becomes x\equiv 5(\bmod 7). The congruence, 3x\equiv 4(\bmod 11) is equivalent to 4(3x)\equiv 4\cdot 4(\bmod 11) which becomes x\equiv 5(\bmod 11).

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

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

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

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

    Combine,
    x\equiv 82(\bmod 385)
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