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Math Help - inverse modulo

  1. #1
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    inverse modulo

    Find the inverse modulo m of each integer n below:

    a) n=8 and m=35

    b) n=51 and m=99
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  2. #2
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    The answer to a is 22.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    The answer to a is 22.

    could u give more details to that??
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by rmpatel5 View Post
    Find the inverse modulo m of each integer n below:

    a) n=8 and m=35
    Find p such that 8p=1 \bmod 35

    Euclidian division :

    35=8 \times 4+3 \implies 3=35-8 \times 4

    8=3 \times 3-1 \implies 1=3 \times 3-8=3 \times (35-8 \times 4)-8=3 \times 35-12 \times 8-8=3 \times 35-13 \times 8

    Therefore (-13) \times 8=1-3 \times 35 =1 \bmod 35

    So the inverse is -13 \bmod 35, that is to say 22 \bmod 35

    b) n=51 and m=99
    Try to do it...
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  5. #5
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    Quote Originally Posted by rmpatel5 View Post
    b) n=51 and m=99
    Okay, there was a trap here...

    You're looking for q such that 51q=1 \bmod 99
    But this is not possible since 51 and 99 are not coprime.

    51q=1 \bmod 99 \Longleftrightarrow \exists k \in \mathbb{Z} ~,~ 51q-1=99k \implies 1=51q-99k=3(17q-33k)

    So 3 must divide 1, which is not possible.
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