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Math Help - Remainders again

  1. #1
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    Remainders again

    find the remainder when 1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998} is divided by 1998

    i tried using eulers theorem and got only this much

    1992^{1980}\equiv 1mod 1998
    similarly for oders...but cudnt get after that

    so i need a kind help.

    thanku
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  2. #2
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    first of all can anyone tell me if this q uses any modular arithmetic..

    bec its getting very untidy with congruency...
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by banku12 View Post
    find the remainder when 1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998} is divided by 1998

    i tried using eulers theorem and got only this much

    1992^{1980}\equiv 1mod 1998
    similarly for oders...but cudnt get after that

    so i need a kind help.

    thanku
    If you got that they are all congruent to one then wouldnt the expression be congruent to -2=1996?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by banku12 View Post
    find the remainder when 1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998} is divided by 1998

    i tried using eulers theorem and got only this much

    1992^{1980}\equiv 1mod 1998
    similarly for oders...but cudnt get after that

    so i need a kind help.

    thanku
     1992^{1980}\equiv 1296\mod{1998}

    Here's a hint:  \phi(1998) = 648 and  1998 = 3\cdot648+54 .
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  5. #5
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    seriously not getting

    can any one of u plz post a complete soln
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    I should have mentioned in my last hint that for Euler's Theorem to work i.e.  a^{\phi(n)}\equiv1\mod{n} , we need  (a,n)=1 . So Euler's Theorem doesn't apply to every term here.

    Try this:

     1992\equiv-6\mod{1998}

    So  1992^{1998}\equiv6^{1998}\mod{1998} .

    Now do something similar to this: http://www.mathhelpforum.com/math-he...remainder.html
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