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Math Help - Modulo airthmetic

  1. #1
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    Modulo airthmetic

    In F41* = (Z/41Z)* determine 17^(-12).

    I deduced that 17^(-12) is congruent to (17^(-1))^12 and congruent to 29^12 (mod 41), since 17*29 = 493 = 12*41 + 1 which is congruent ti 1 (mod 41)

    After that I can't find a way to reduce 29^12 (modulo 41)

    Any help would be appreciated!!!
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  2. #2
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    Well 29^2=841 which is congruent to 21. So 29^{12}=(29^2)^6=21^6. Can you take it from here?
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  3. #3
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    Thanks for the help! 21^6 congruent to (21^2)^3 congruent to (-10)^3 but then? I mean we can't take (21^3)^2 since 21^3 is a large number. I really can't think right now any other possibility.
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  4. #4
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    21^2=441 congruent to 31. 31^2=961 congruent to 18. 18*31=558 congruent to 25. So, 21^6=(21^2)^3=31^3=31^2*31=18*31=25.
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  5. #5
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    Thank you very much for your help. Appreciate it!!
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