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anonymous

  • one year ago

solve the following system. \[2x\equiv 3 (mod 5)\] \[4x\equiv 2 (mod 6)\] \[3x\equiv 2 (mod 7)\]

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  1. ganeshie8
    • one year ago
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    familiar with chinese remainder theorem ?

  2. IrishBoy123
    • one year ago
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    .

  3. anonymous
    • one year ago
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    I have done an assignment but it only had something like \[x\equiv 3 (mod 5)\] \[x\equiv 2 (mod 6)\] \[x\equiv 2 (mod 7)\]. But am not sure how to use it to solve the problem. @ganeshie8

  4. ganeshie8
    • one year ago
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    solve each of the linear congruence to get a system with simpler congruences, for example : \(2x\equiv 3 \pmod 5\) simplifies to \(x\equiv 4\pmod{5}\)

  5. anonymous
    • one year ago
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    @ganeshie8 Please show me a step on how you got the 4

  6. ganeshie8
    • one year ago
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    I have just guessed it. Plugin x=4 and observe that it satisfies the congruence \[2*4\equiv 8\equiv 3\pmod{5}\]

  7. anonymous
    • one year ago
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    @ganeshie8 Thanks. I will try doing that and then apply the Chinese remainder theorem

  8. ganeshie8
    • one year ago
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    For the second congruence, \(4x\equiv 2 \pmod 6\), dividing \(2\) through out gives \[2x\equiv 1\pmod{3}\] then it is easy to eyeball the solution : \[x\equiv 2\pmod{3}\]

  9. anonymous
    • one year ago
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    \[3x\equiv 2 (mod 7)\] simplifues to \[x\equiv -4 (mod 7)\] Does that look correct? @ganeshie8

  10. anonymous
    • one year ago
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    Probably \[x\equiv 10 (mod 7)\]

  11. ganeshie8
    • one year ago
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    \(x\equiv -4\pmod{7}\) is same as \(x\equiv 3 \pmod{7}\)

  12. ganeshie8
    • one year ago
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    so the given system, after transforming into simple linear congruences is \[x\equiv 4\pmod{5}\\x\equiv 2\pmod{3}\\x\equiv 3\pmod{7}\]

  13. anonymous
    • one year ago
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    I got it. Thanks alot @ganeshie8

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