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Math Help - Modulus functions

  1. #1
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    Modulus functions

    Hi, I've got so far try to solve simultaneous congruence equations but have run a ground.

    Question is solve the simultaneous equations

    7x = 3mod34 and x=7mod41


    I've solved the first one to get x= 15mod 34 hence x= 34u+15

    The second is x= 7 mod 41

    Thus I've put 34u+15 = 7mod 41

    Now at this point I thought it would go to 34u = 33mod41 but if this is the case then I'm struggling to solve for u. I think I've made a mistake at 34u+15 = 7 mod 41 but not sure what.
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  2. #2
    Senior Member roninpro's Avatar
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    I'm not sure what the problem is. You have 34u\equiv 33\pmod{41}. Multiply both sides by 5 to find u\equiv 15\pmod{41}. Then, u=41v+15 and x=34(41v+15)+15=1394v+525. In summary, any solution to the system has the form x\equiv 525\pmod{1394}.
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  3. #3
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    I'm a bit confused on the middle step, how did you get to u=15mod41?

    I don't understand how you figured out to multiply it by 5.
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  4. #4
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    FYI, the Chinese Remainder Theorem gives a method for solving a set of congruences of the form x\equiv{a_i}\ (\text{mod }m_i), where the m_i are pairwise relatively prime.

    Your solution is probably just as easy though.

    The number 5 that roninpro used should be the inverse (mod 41) of 34. In other words, it is the solution to 34x\equiv{1}\ (\text{mod }41). I believe this is 35, not 5. 5 is the inverse of 33, not 34. The Extended Euclidean Algorithm can find this inverse if the numbers are larger and brute force is too hard.

    So u=35*33=7 (mod 41), and x=34(41v+7)+15=1394v+253, or x=253 (mod 1394).

    As a check,
    7*253=1771=3 (mod 34)
    253=7 (mod 41)

    - Hollywood
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