Multiplicative Inverse using Fermats theorm Which of the following is a multiplicative inverse of $11^{23}$ modulo $59$?


*

*$11^{21}$

*$11^{22}$

*$11^{25}$

*$11^{35}$

*$11^{60}$


I assume that I'm supposed to use Fermat's little theorem in order to show $11^{58}\equiv1\pmod{59}$.
And from there I could probably say that $11^58$ is equal to $11^{29}\times2$, so that's also an inverse.
But I can't see how I get to any of the answers listed.
 A: You're on the right track:
$$
1 \equiv 11^{58} = 11^{23} 11^{35} \bmod 59
$$
A: Since
$$11^{58}=1\pmod{59}\;,\;\;\text{ you need ...}$$
A: ${\rm mod}\ 59\!:\ \overbrace{\color{#c00}{11^{\large k}}\times \color{#0a0}{11^{\large 58-k}}}^{\quad \Large 11^{58}}\!\equiv 1\ $ by lil Fermat.
Thus $\,(\color{#c00}{11^{\large k}})^{-1}\! \equiv \color{#0a0}{11^{\large 58-k}}\,$ by $ $ uniqueness of inverses.
A: Let $p$ be prime and $x\not \equiv 0\pmod p.$
(1). Let $o(x)$ be the least $n\in \mathbb N$ such that $x^n\equiv 1\pmod p.$ Then $o(x)\;|\;( p-1).$ 
Proof:   Let $d=\gcd (o(x),p-1).$ There are integers $A,B$ with $Ao(x)+B(p-1)=d.$ $$\text {So }\quad x^d=(x^{o(x)})^A(x^{p-1})^B\equiv 1^A1^B\equiv 1 \pmod p.$$ Therefore, by def'n of $o(x)$ we have $d\geq o(x)$. But by def'n of $d$ we have $d\leq o(x).$ 
(2). If  $n\ne 0$ and $x^n\equiv 1\pmod p$ then $o(x)\;|\;n.$
Proof: Let $e=\gcd (n,o(x)).$ There are integers $E,F$ with $En+Fo(x)=e.$ $$\text {So }\quad  x^e=(x^n)^E(x^{o(x)})^F\equiv 1^E1^F\equiv 1 \pmod p.$$  Therefore, by def'n of $o(x)$ we have  $e\geq o(x)$. But by def'n of $e$ we have $o(x)\leq e.$ 
(3). With $p=59$  we have $o(11)\;|\;58=2\cdot 29.\;$ So $o(11)\in \{1,2,29,58\}.$  Since $11^2\not \equiv 1\not \equiv 11^1  \pmod {59}$ we have $o(11)\in \{29,58\}.$ So if $11^n\equiv 1 \pmod {59}$ then $29|n.$
Whether $o(11)=29$ or $o(11)=58$ is not needed for your Q.
