# Math Help - Modulo airthmetic

1. ## 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!!!

2. Well $29^2=841$ which is congruent to $21$. So $29^{12}=(29^2)^6=21^6$. Can you take it from here?

3. 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.

4. $21^2=441$ congruent to $31$. $31^2=961$ congruent to $18$. $18*31=558$ congruent to $25$. So, $21^6=(21^2)^3=31^3=31^2*31=18*31=25$.

5. Thank you very much for your help. Appreciate it!!