This is something I have been thinking of using in a math competition against other players so it would be very helpful to me if it was explained.
How would someone reduce a problem such as $\frac{7^{54}}{11}$ mod 11 or any problem in the form $\frac{a^b}{c} \bmod \{7,8,9,10,11\} $ with the Carmichael function?
In the problem I've listed, $\lambda(11^2)=110$ but I do not know where to go from here.
I know that half of 110 is 55, and the exponent 54 is one less than 55 . But I'm not sure how to apply this piece of information.
Thank you for helping.
So if we want to find for example $$x\equiv\frac{7^{54}}{13} \mod{11}$$ $$13x\equiv7^{54} \mod{11}$$ Now because $\lambda(11)=10$ and $\gcd{(7,11)}=1$ we now have that $$7^{10}\equiv1\mod{11}$$ $$\therefore 7^{54}\equiv7^4\equiv49^2\equiv5^2\equiv25\equiv3\mod{11}$$ So we now have to solve $$13x\equiv3\mod{11}$$ which has solutions $$x\equiv7\mod{11}$$