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anonymous

  • one year ago

solve 3x ≡ 1 (mod 5) 4x ≡ 6 (mod 14) 5x ≡ 11 (mod 3)

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  1. anonymous
    • one year ago
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    I found x = 82 or x = 187 but what are ALL solutions?

  2. ganeshie8
    • one year ago
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    did you use chinese remainder thm?

  3. anonymous
    • one year ago
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    I think I did.

  4. anonymous
    • one year ago
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    It only said Theorem but didn't explicitly said "Chinese Remainder Theorem" Let me post the link

  5. ganeshie8
    • one year ago
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    chinese remainder thm gives you the solution \[x\equiv 82\pmod{5*7*3}\]

  6. beginnersmind
    • one year ago
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    You can divide eq1 by 3 and eq3 by 5, since they are moduli over primes.

  7. anonymous
    • one year ago
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    here: https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA58&pg=PA68#v=onepage&q&f=false

  8. anonymous
    • one year ago
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    Oh so I was supposed to divide the second congruence by 2?

  9. ganeshie8
    • one year ago
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    that is chinese remainder theorem

  10. ganeshie8
    • one year ago
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    it doesn't affect the final solution dividing by 2 simplifies the work, thats all

  11. beginnersmind
    • one year ago
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    I don't think you can divide eq2. Actually division is the wrong word.

  12. anonymous
    • one year ago
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    what about the solution 187 (mod 5*14*3) ??

  13. ganeshie8
    • one year ago
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    in mod 5*14*3, you should get "two" incongruent solutions

  14. ganeshie8
    • one year ago
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    ``` I found x = 82 or x = 187 but what are ALL solutions? ``` Yes, those are all the incongruent solutions in mod 5*14*3

  15. anonymous
    • one year ago
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    oh so in a sense {82 + 5*7*3k, k integer} is equal to the set {82 + 5*14*3k} U {187 + 5*14*3k} ??

  16. beginnersmind
    • one year ago
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    Lets take equation 1 as an example 3x ≡ 1 (mod 5) says that if you divide 3x by 5 you get a remainder of one. We want to know x (mod 5) is. Since 5 is a prime number we can determine the remainder of x:5 uniquely. In particular x ≡ 2 (mod 5). Similarly for equation three 5x ≡ 11 (mod 3) -> 5x ≡ 2 (mod 3) -> x ≡ 1 (mod 3)

  17. beginnersmind
    • one year ago
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    Since 14 is not a prime number 4x ≡ 6 (mod 14) doesn't have a unique solution for x (mod 14)

  18. anonymous
    • one year ago
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    I see. Thank you both :')

  19. beginnersmind
    • one year ago
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    So the next step is to find all x (mod 14) that solve 4x ≡ 6 (mod 14).

  20. ganeshie8
    • one year ago
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    ``` oh so in a sense {82 + 5*7*3k, k integer} is equal to the set {82 + 5*14*3k} U {187 + 5*14*3k} ?? ``` Precisely @sourwing !

  21. anonymous
    • one year ago
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    @ganeshie8 thank you!!

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