Solve $ord_x(2) = 20$ Given that the (multiplicative) order of $2$ mod $x$ is $20$, how can I work out what $x$ is?

You need $$2^k\cdot2^{20-k}=1$$ for $k=1,2,...,20$. In other words $2^k$ has as inverse $2^{20-k}$ Since $2$ is a primitive root modulo $11$ we examine if this is verified for $x=11$. We have in the field $\mathbb F_{11}$ $$\begin{cases}2^1=2\\2^2=4\\2^3=8\\2^4=5\\2^5=10\\2^6=9\\2^7=7\\2^8=3\\2^9=6\\2^{10}=1\end{cases}$$ and we can see the condition $2^k\cdot2^{20-k}=1$ is verified. So $\color{red}{x=11}$
Is there another solution? I will come back (some trouble to write in English and I could less the first answer as many times).
NOTE.-This is an answer to the question as it was formulated before it was changed: this question was $2^{20}\equiv 1\pmod x$
