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Math Help - Remainder problem

  1. #1
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    Remainder problem

    Q) Find the remainder obtained when 43^101 + 23^101 is divided by 66?

    Thanks,
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by a69356 View Post
    Q) Find the remainder obtained when 43^101 + 23^101 is divided by 66?

    Thanks,
    Consider the remainder when divided by 11 and the remainder when divided by 6.

    That is look at:

    43^{101}=(11\times 4-1)^{101} = k\times 11-1

    for some k\in \mathbb{N}

    Now do something similar for 23 ..

    (MOdular arithmetic to base 11 and 6 will also work if you are familiar with it)

    CB
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  3. #3
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    Altought we can use the relation

    <br />
N^{s+1}\equiv N \quad mod \quad Denominator [B_s]<br />
    Bs is the s-th Bernoulli Number
    (100-th Bernoulli Denominator is 66 \cdot 505)

    (Look Bernoulli Numbers Fundamental Theorem of Arithmetic (in google) for details),

    A simpler approach is:

    <br />
43^{101}+23^{101}=(43+23)(43^{100}-43^{99}23+....-43\cdot23^{99}+23^{100})<br />

    Look here for this factorization

    The last expression obviously is divisible by 66, so:

    <br />
43^{101}+23^{101}\equiv 0 \quad mod\quad 66<br />
    Last edited by gdmath; August 17th 2009 at 02:06 AM.
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