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Math Help - congruence problem

  1. #1
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    congruence problem

    suppose b is congruent to a^67 (mod 91) and (a,91)=1 (gcd) for some a,b element of Z such that 0<=a<91.

    1) find k element of Z such that b^k is congruent to a (mod 91)
    2) let b=53. Find a

    i think 1 & 2 need to be done as separate questions but I have NO clue how to do this.

    any help is appreciated.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Are we doing all your homework tonight?

    1)
    b \equiv a^{67} \mod 91
    b^k \equiv a^{67k} \mod 91

    So you want to find k such that 67k \equiv 1 \mod 91; i.e. you want to find the inverse of 67 in the group (\mathbb{Z}/91\mathbb{Z})^\times. To do this, note that for any x \in (\mathbb{Z}/91\mathbb{Z})^\times, x^{\phi(91)}=1; hence 67^{\phi(91)}=1, or 67(67^{\phi(91)-1})=1... can you finish now?
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  3. #3
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    no, i have no idea how to finish.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Bruno J. View Post
    So you want to find k such that 67k \equiv 1 \mod 91
    Actually, you have \varphi(91)=72 (where \varphi is the Euler function) so a^{72}\equiv1\,(\bmod\,91).

    So you want to find k such that 67k\equiv1\,(\bmod\,\color{red}72\color{black}) (not mod 91).
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