Calculate the Following
$ (2^{19808}+6)^{-1} +1$ Mod (11)
I'm completely lost here for several reasons. First of all the large power of 2 just throws me off and secondly I've seen inverse equations before never an inverse + a number.
Could someone explain what's going on please?
In order to compute $$x \equiv (2^{19808}+6)^{-1} \pmod {11}$$ we have to solve $$(2^{19808}+6)x\equiv 1 \pmod {11}$$ By Euler's theorem we have $$2^{19808} \pmod {11} \equiv 2^{19808 \bmod \varphi(11)} \pmod {11} \equiv 2^{19808 \bmod 10} \pmod {11}$$ $$ \equiv 2^8 \pmod {11} \equiv 256 \pmod {11} \equiv 3 \pmod {11}$$ and therfore $$(3+6)x\equiv 9x \equiv1 \pmod {11}$$ Thus $x\equiv 5 \pmod {11}$ and finally $$ (2^{19808}+6)^{-1} +1 \equiv 5 + 1 \equiv 6 \pmod {11}$$
Another approach, perhaps a little shorter and/or easier to follow: since $$19808=8\begin{cases}\pmod{10}\\\pmod{11}\end{cases} \;\text{and}\;\;2^{10}=1\pmod{11}\,,\;\text{ (Fermat's Little Theorem), we get that }$$
(in the following all is done modulo $\;11\;$) :
$${}$$
$$2^{19808}=2^8=2^5\cdot 2^3=-2^3=-8=3$$
so
$$2^{19808}+6=3+6=9\;,\;\;\text{and since}\;\;9\cdot 5=1$$
we have that
$$\left(2^{19808}+6\right)^{-1}+1=9^{-1}+1=5+1=6$$
