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Math Help - Help Preparing for Final Exam

  1. #16
    Member diddledabble's Avatar
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    New Question on Congrunences

    Solve the congruence given 2x=1(mod 3), x=3(mod 5), x=2(mod 8)
    Here is what I did.

    2x=4(mod 3)
    x=2(mod 3)
    x=2+3n

    2+3n=3(mod 5)
    3n=1(mod 5)
    3n=6(mod 5)
    n=2(mod 5)
    then n=2+5k

    x=2+3(2+5k)
    x=2+6+15k
    x=8(mod 15)

    Let x=8+15k
    8+15k=2(mod 8)
    15k=-6(mod 8)
    15k=4(mod 8)
    15k=60(mod 8)
    k=4(mod 8)
    k=4+8m

    x=8(mod 15)
    x=8+15(4+8m)
    x=8+15(4+8m)
    x=8+60+120m
    x=68+120m
    x=68(mod 120)

    Did I get that right?
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  2. #17
    Super Member Gamma's Avatar
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    Iowa City, IA
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    Quote Originally Posted by diddledabble View Post
    Solve the congruence given 2x=1(mod 3), x=3(mod 5), x=2(mod 8)
    Here is what I did.

    2x=4(mod 3)
    x=2(mod 3)
    x=2+3n

    2+3n=3(mod 5)
    3n=1(mod 5)
    3n=6(mod 5)
    n=2(mod 5)
    then n=2+5k

    x=2+3(2+5k)
    x=2+6+15k
    x=8(mod 15)

    Let x=8+15k
    8+15k=2(mod 8)
    15k=-6(mod 8)
    15k=4(mod 8)
    15k=60(mod 8)
    k=4(mod 8)
    k=4+8m

    x=8(mod 15)
    x=8+15(4+8m)
    x=8+15(4+8m)
    x=8+60+120m
    x=68+120m
    x=68(mod 120)

    Did I get that right?
    Start a new thread when you have a new question.

    No. 68\equiv 4 \not \equiv 2 (mod 8)
    Last edited by Gamma; August 4th 2009 at 12:43 PM. Reason: added 2 for clarity
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  3. #18
    Super Member Gamma's Avatar
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    Iowa City, IA
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    Quote Originally Posted by diddledabble View Post
    Solve the congruence given 2x=1(mod 3), x=3(mod 5), x=2(mod 8)
    Here is what I did.

    2x=4(mod 3)
    x=2(mod 3)
    x=2+3n

    2+3n=3(mod 5)
    3n=1(mod 5)
    3n=6(mod 5)
    n=2(mod 5)
    then n=2+5k

    x=2+3(2+5k)
    x=2+6+15k
    x=8(mod 15)

    Let x=8+15k
    8+15k=2(mod 8)
    15k=-6(mod 8)
    15k=4(mod 8)
    15k=60(mod 8)
    k=4(mod 8)
    k=4+8m

    x=8(mod 15)
    x=8+15(4+8m)
    x=8+15(4+8m)
    x=8+60+120m
    x=68+120m
    x=68(mod 120)

    Did I get that right?
    There is your error. -6\equiv 2 (mod 8) not 4.

    But you need to be careful dividing by stuff in there man, that is a recipe for disaster. There are zero divisors and things without multiplicative inverses all over the place in the integers mod 8. In fact in any modulus that is not prime.

    98 (mod 120) is what you are looking for.

    I suggest looking into the chinese remainder theorem to solve these problems. http://www.mathhelpforum.com/math-he...r-theorem.html
    Last edited by Gamma; August 4th 2009 at 01:04 PM. Reason: added reference
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  4. #19
    Member diddledabble's Avatar
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    Post Where did I go wrong?

    I would repost but don't want to get in trouble for double posting this so I will leave it here, but in the future I will make a new thread. Thanks Gamma.

    In the meantime can you point out where my logic went wrong? In the congruence.
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  5. #20
    Super Member Gamma's Avatar
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    Iowa City, IA
    Posts
    517
    If it is a different question it is not double posting. What people do not want to see is some guy asking a question and if no one answers it within an hour they repost again to get theirs to the top. Then no one answers it still and they repost again with a title HELP ME NOW!!!! URGENT OMG NEED YOUR HELP.

    It is important to make each question in a different thread with a descriptive title so people browsing can quickly decide if they are going to know how to help you or not. No one will know that you have 3 other questions hidden inside some 17 post thread about gcds that was answered already in the first response.

    I pointed out your error above.
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