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Math Help - 9^103

  1. #1
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    9^103

    Find the sum of the last 2 digits of 9^103

    What is the product of the last 2 digits of 2004^2004

    What other ways to do these besides Euler's theorem and finding the pattern?

    if you use factorials for (a-b)^103 you can write the equation back to front and add it together to get twice the value but cancelling out the intermediate factorials then divde the equation by half to get the number itself then just add the last two digits of it.

    Someone explain it?
    Last edited by masterwizard; February 20th 2009 at 03:00 AM.
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  2. #2
    MHF Contributor

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    Use the binomial theorem.

    9^{103}= (10- 1)^{103}. By the binomial theorem, the last two terms of (x- y)^{103} are \left(\begin{array}{c}103 \\ 102\end{array}\right)xy^{102}+ \left(\begin{array}{c}103 \\ 103\end{array}\right)y^{103}= 103xy^{102}+ y^{103}. The last two terms of this particular binomial expansion are 103(10)(1)+ (-1)= 102. All higher terms involve integers times 100 or higher so the last two digits are "02".

    Similarly 2004^{2004}= (2000+ 4)^{2004}. Now the last two terms are 2004(2000)4^{2003}+ 4^{2004} so the problem reduces to finding the last two digits of 4^{2004}.
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  3. #3
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    How do I calculate 4^2004 using the same method? How do I break the 4 down? If I break it down to 3+1 or 5-1, there's a 1 there. And multiples of 4 don't have 1 as their last digit
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