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Math Help - Cryptography

  1. #1
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    Cryptography

    I was also wondering if someone could help me find the answers to these simple problems

    -16 mod 3

    -52 mod 19

    is there a formula to follow for these problems?



    Could someone help me get a solution to this probelm.

    Compute 2^{1000000} mod 7

    Repeated Square method might do it.
    Last edited by flinted; May 19th 2008 at 08:15 AM.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by flinted View Post
    Could someone help me get a solution to this probelm.

    Compute 2(^1000000) mod 7
    2^3=8=1 \mod 7

    1000000=3k+1

    Therefore, 2^{1000000}=2^{3k+1}=2 \cdot (2^3)^k \equiv 2 \cdot 1^k \mod 7

    \boxed{2^{1000000} = 2 \mod 7}
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  3. #3
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    Quote Originally Posted by flinted View Post
    Could someone help me get a solution to this probelm.

    Compute 2(^)((1000000)) mod 7
    To find 2^{1000000} mod 7, observe that 2^3 = 1 mod 7. This means 2^{999999} = (2^3)^{333333} = 1^{333333} mod 7. 2^{999999} = 1 mod 7 \Rightarrow 2^{1000000} = 2 mod 7
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  4. #4
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    thanks but i dont undertand where you got the k + 1 formula from
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  5. #5
    Moo
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    1000000=3x333333+1
    I substitute 333333 by k.
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  6. #6
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    thanks alot
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  7. #7
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    *bump*
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