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Math Help - Remainder when large numbers divided

  1. #1
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    Remainder when large numbers divided

    What is the remainder when 2001^{2001} is divided by 26?

    I think I have to do something with 26, like break it up into 13 * 2. Try to use Fermat's Theorem somehow. I know that (2001, 13) = 1 and (2001, 2) = 1.

    Any help to get rolling would be appreciated.
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  2. #2
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    Re: Remainder when large numbers divided

    2001 \equiv 25 \pmod {26}
    2001^2 \equiv 25^2 \equiv 1 \pmod {26}
    2001^4 \equiv 1^2 \equiv 1 \pmod {26}
    2001^8 \equiv 1 \pmod {26}
    2001^{16} \equiv 1 \pmod {26}
    2001^{64} \equiv 1 \pmod {26}
    2001^{256} \equiv 1 \pmod {26}
    2001^{512} \equiv 1 \pmod {26}
    2001^{1024} \equiv 1 \pmod {26}

    2001^{2001} = 2001^{1024} 2001^{512} 2001^{256} 2001^{128} 2001^{64} 2001^{16} 2001 \equiv 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 25 \equiv 25 \pmod {26}
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  3. #3
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    Re: Remainder when large numbers divided

    Hello, Zalren!

    \text{What is the remainder when }2001^{2001}\text{ is divided by }26\,?
    . . . 2001 \;=\;77(26) -1

    \begin{array}{ccccc} 2001 & \equiv & \text{-}1 & \text{(mod 26)} \\ \\[-3mm] 2001^{2001} & \equiv & (\text{-}1)^{2001} & \text{(mod 26)} \\ \\[-3mm] 2001^{2001} & \equiv & \text{-}1 & \text{(mod 26)} \\ \\[-3mm] 2001^{2001} & \equiv & 25 & \text{(mod 26)}\end{array}


    \text{Therefore, }2001^{2001} \div 26\text{ has a remainder of }25.

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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Remainder when large numbers divided

    Quote Originally Posted by VinceW View Post
    2001 \equiv 25 \pmod {26}
    2001^2 \equiv 25^2 \equiv 1 \pmod {26}
    2001^4 \equiv 1^2 \equiv 1 \pmod {26}
    2001^8 \equiv 1 \pmod {26}
    2001^{16} \equiv 1 \pmod {26}
    2001^{64} \equiv 1 \pmod {26}
    2001^{256} \equiv 1 \pmod {26}
    2001^{512} \equiv 1 \pmod {26}
    2001^{1024} \equiv 1 \pmod {26}

    2001^{2001} = 2001^{1024} 2001^{512} 2001^{256} 2001^{128} 2001^{64} 2001^{16} 2001 \equiv 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 25 \equiv 25 \pmod {26}
    I think you made it harder than it has to be (both of you). Isn't it true that \displaystyle 2001^{2001}\equiv 25^{2001}\equiv (-1)^{2001}\equiv -1\equiv 25\text{ mod }26.
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