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  1. #1
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    help meeee

    9^2007 devided by 8. find the remainder
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    Quote Originally Posted by bananalion View Post
    9^2007 devided by 8. find the remainder
    Are you doing modular mathematics?

    9 = 1 (mod 8)

    So
    9^n = 1^n (mod 8) = 1 (mod 8) for integer n

    Thus 9^{2007} = 1 (mod 8) so 9^{2007}/8 has a remainder of 1.

    -Dan
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    no......the question is only on the algebra.....u think its possible?
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    Quote Originally Posted by bananalion View Post
    no......the question is only on the algebra.....u think its possible?
    Just look at the pattern then.

    9 leaves remainder 1 upon division by 8.

    9^2 = 81 leaves remainder 1 upon division by 8.

    9^3 =729 leaves reaminder 1 upon division by 8.

    It seems that 9 to any exponenets leaves the same remainder.
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  5. #5
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    thank you very much

    then bout what about this question?

    2003x2004x2005+2006^3 devided by 7

    but not using calculator
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    do you think the remainder is 0?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by bananalion View Post
    thank you very much

    then bout what about this question?

    2003x2004x2005+2006^3 devided by 7

    but not using calculator
    2003*2004*2005 = (2006 - 3)(2006 - 2)(2006 - 1)

    = 2006^3 - 6*2006^2 + 11*2006 - 6

    So
    2003*2004*2005 + 2006^3 = 2*2006^3 - 6*2006^2 + 11*2006 - 6

    Now, similar to the idea from before, let's get the remainders of each of the powers of 2006:
    2006/7 leaves a remainder of 4
    2006^2/7 leaves a remainder of 4^2 = 16
    2006^3/7 leaves a remainder of 4^3 = 64

    Now,
    16 = 2*7 + 2, so this is a remainder of 2.
    64 = 9*7 + 1, so this is a remainder of 1.

    Thus
    (2003*2004*2005 + 2006^3)/7 = (2*2006^3 - 6*2006^2 + 11*2006 - 6)/7

    leaves a remainder of:
    2*1 - 6*2 + 11*4 - 6 = 2 - 12 + 44 - 6 = 28

    And 28 = 4*7 + 0

    So you are correct, this leaves a remainder of 0 when you divide by 7.

    -Dan
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