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Math Help - Is there a quicker way to solve this congruence

  1. #1
    Super Member Deadstar's Avatar
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    Is there a quicker way to solve this congruence

    Is 2^{35} \equiv 1 \textrm{ mod } 561?

    In the question we were given maple commands to use so I assume we were meant to do it that way. Is there a way to solve this without a calculator?

    I had to use one but just did it like...

    2^8 = 256,
    so 2^{16} = 450 mod 561
    and 2^{32} = (2^{16})^2 = 103 mod 561...

    Then 2^{35} = 2^{32} 2^3 = 103 \times 8 which is not equal to 1 mod 561..
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Is 2^{35} \equiv 1 \textrm{ mod } 561?


    Since 561=3\cdot 11\cdot 17 , we can do:

    *** Arithmetic modulo 3 : 2^{35}=(2^3)^{11}\cdot 2^2=2^{11}\cdot 2^2=(2^3)^4\cdot 2=2^4\cdot 2=2^3\cdot 2^2=2^3=2\!\!\!\pmod 3

    *** Arithmetic modulo 11 : 2^{35}=(2^{11})^3\cdot 2^2=2^3\cdot 2^2=32=10=-1\!\!\!\pmod {11}

    *** Arithmetic modulo 17 : 2^{35}=(2^{17})^2\cdot 2=2^2\cdot 2=8\!\!\!\pmod {17} .


    Well, now use the Chinese Remainder Theorem to find an element x\in\mathbb{Z}\,\,\,s.t.\,\,\,x=2\!\!\!\pmod 3\,,\,x=-1\!\!\!\pmod {11}\,,\,x=8\!\!\!\pmod{17} .

    I found x=-298=263\!\!\!\pmod{561}\Longrightarrow 2^{35}=263\!\!\!\pmod{561} , as you can easily check with any calculator.

    Tonio






    In the question we were given maple commands to use so I assume we were meant to do it that way. Is there a way to solve this without a calculator?

    I had to use one but just did it like...

    2^8 = 256,
    so 2^{16} = 450 mod 561
    and 2^{32} = (2^{16})^2 = 103 mod 561...

    Then 2^{35} = 2^{32} 2^3 = 103 \times 8 which is not equal to 1 mod 561..
    .
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