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anonymous

  • 4 years ago

Find all solutions of the following linear congruence. \(3x\equiv2\mod7\) First of all, we notice that \((3,7)=1\). Therefore, we will only have \(1\) solution. We now need to obtain a solution of the linear, diophantine equation \(3x-7y=2\). The Euclidean algorithm gives: \(7=3\cdot2+1\), \(3=1\cdot3+0\). Hence, \(7\cdot1-3\cdot2=1\) and \(7\cdot2-3\cdot4=2\). Therefore, a particular solution to the diophantine equation is \(x_0=-4\), \(y_0=-2\) and all solutions of the linear congruences are given by \(x\equiv-4\equiv3\mod7\). Am I right?

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  1. myininaya
    • 4 years ago
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    I have a way but you might now like it

  2. myininaya
    • 4 years ago
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    not*

  3. anonymous
    • 4 years ago
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    i am sure this is right, but i would write the following \[3x\equiv2\mod7\] \[3x\equiv9\mod7\] \[x\equiv3\mod7\]

  4. myininaya
    • 4 years ago
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    \[3x \equiv 2 \mod 7 =>3x=2+7k\] k=0, there is no integer x k=1, x=3 k=2, there is no integer x k=3, there is no integer x k=4, x=10 k=5, there is no integer x k=6, there is no integer x k=7, x=17 so on... so we have x=...,3,10,17,17+7,(17+7)+7, and so on...

  5. anonymous
    • 4 years ago
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    that is the usual gimmick. keep adding until you can divide to solve the congruence

  6. myininaya
    • 4 years ago
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    x=3+7i where i is an integer

  7. myininaya
    • 4 years ago
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    which is what satellite said

  8. anonymous
    • 4 years ago
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    in two lines lol

  9. myininaya
    • 4 years ago
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    shhh... don't make fun of my inefficient way ;)

  10. anonymous
    • 4 years ago
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    speed kills

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