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Math Help - Modulo Stuff...

  1. #1
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    Modulo Stuff...

    '=' means 'congruent to'

    (Uses Chinese Remainder Theorem)

    x=0(mod4) --> 3600*Y1=1(mod4)
    x=1(mod9) --> 1600*Y2=1(mod9)
    x=2(mod16) --> 900*Y3=1(mod16)
    x=3(mod25) --> 576*Y4=1(mod25)

    Knowing that M1=3600, M2=1600, M3=900, and M4=576

    Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

    Any pointers will do.
    Thanks!
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  2. #2
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    San Diego
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    Quote Originally Posted by AgentNXP View Post
    '=' means 'congruent to'

    (Uses Chinese Remainder Theorem)

    x=0(mod4) --> 3600*Y1=1(mod4)
    x=1(mod9) --> 1600*Y2=1(mod9)
    x=2(mod16) --> 900*Y3=1(mod16)
    x=3(mod25) --> 576*Y4=1(mod25)

    Knowing that M1=3600, M2=1600, M3=900, and M4=576

    Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

    Any pointers will do.
    Thanks!
    There is a problem because you can't have x=0(mod4) and x=2(mod16) in the same system.
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  3. #3
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    Quote Originally Posted by AgentNXP View Post
    '=' means 'congruent to'

    (Uses Chinese Remainder Theorem)

    x=0(mod4) --> 11025*Y1=1(mod4)
    x=1(mod9) --> 4900*Y2=1(mod9)
    x=2(mod25) --> 1764*Y3=1(mod25)
    x=3(mod49) --> 900*Y4=1(mod49)

    Knowing that M1=11025, M2=4900, M3=1764, and M4=900

    Is it possible to 'simplify' the congruences to the right? I know I couldn't... or is there a different approach to solving the system of congruences to the left?

    Any pointers will do.
    Thanks!
    Crap! Thanks for noticing that. Fixed above.

    I'll try it again now--cheers!
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  4. #4
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    x=0(mod4) --> 11025*Y1=1(mod4) --> Y1=1(mod4)
    x=1(mod9) --> 4900*Y2=1(mod9) --> Y2=7(mod9)
    x=2(mod25) --> 1764*Y3=1(mod25) --> Y3=9(mod25
    x=3(mod49) --> 900*Y4=1(mod49) --> Y4=43(mod49)

    Knowing that M1=11025, M2=4900, M3=1764, and M4=900

    So after re-doing the problem...

    I end up getting x=39053(mod 44100)

    Now, what I am trying to get: Four consecutive numbers none of which is “square-free” (that is, each of which can be divided by a perfect square). It's not working so well for me.
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