By what rule or property this expression is equal to six? $$2^{-1} \equiv 6\mod{11}$$
Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?
$$6^{-1} \equiv 2\mod11$$
 A: It's because $2\cdot 6 \equiv 1 \mod 11$
A: First, we need to understand what $2^{-1}\mod 11$ stands for. Raising to the power $-1$ is usually used to denote the multiplicative inverse. That is, $b$ is a multiplicative inverse of $a$ if and only if $a\cdot b=1$. Thus, $2^{-1}\mod 11$ denotes the number such that $$2^{-1}\cdot2\equiv1\mod11$$ Since $2\cdot 6\equiv 12\equiv1\mod11$, we see $2^{-1}\equiv 6\mod11$. We can do the same to show $6^{-1}\equiv 2\mod 11$.

For a reliable approach (that is, one that doesn't rely on simply testing $2\cdot 1,2\cdot 2,\cdots,2\cdot 10$ and see which one is $1$), we can use (where we're using that $\gcd(2,11)=1$) $$2^{\phi(11)}\equiv 1\mod 11$$ by Euler's Totient Function (or you could use Fermat's Little Theorem in this case, since $11$ is prime), and since $\phi(11)=10$, we have \begin{align}
2^{-1}&\equiv 2^{-1}\cdot 1\\
&\equiv 2^{-1}\cdot2^{\phi(11)}\\
&\equiv 2^{-1}\cdot2^{10}\\
&\equiv 2^{-1+10}\\
&\equiv 2^9\\
&\equiv 512\\
&\equiv 6\mod 11
\end{align}
A: You have $2\times 6=12$ which is $1$ mod $11$. Hence, we see that that $2\times 6=1$ mod $11$.
Now the set of congruence numbers modulo $11$ (often noted $\mathbb{Z}/11$) is a ring with addition and multiplication coming from $\mathbb{Z}$. 
Saying that $6^{-1}\text{  mod }11=2$ is saying that $6$ mod $11$ is invertible as an element of the ring $(\mathbb{Z}/11,+,\times)$ and that its inverse is $2$ mod $11$. 
A: The notation $2^{-1} \pmod{11}$ means the multiplicative inverse of $2$ modulo $11$, that is, a number $x$ satisfying the equation $2x \equiv 1 \pmod{11}$.  
Since $2 \cdot 6 \equiv 12 \equiv 1 \pmod{11}$, $2^{-1} \equiv 6 \pmod{11}$.  The same calculation shows that $6^{-1} \equiv 2 \pmod{11}$.  
The multiplicative inverse of $6$ exists modulo $11$ since $\gcd(6, 11) = 1$.
We can find the multiplicative inverse of $6$ modulo $11$ using the extended Euclidean algorithm.
\begin{align*}
11 & = 1 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align*}
We solve for $1$ as a linear combination of $6$ and $11$.
\begin{align*}
1 & = 6 - 5\\
  & = 6 - (11 - 6)\\
  & = 2 \cdot 6 - 11
\end{align*}
Hence, $2 \cdot 6 \equiv 1 \pmod{11} \iff 6^{-1} \equiv 2 \pmod{11}$.
Let's see what happens when we use the extended Euclidean algorithm to solve for $2^{-1}$ modulo $11$.
The multiplicative inverse of $2$ exists modulo $11$ since $\gcd(2, 11) = 1$.
\begin{align*}
11 & = 5 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Solving for $1$ as a linear combination of $2$ and $11$ yields
$$1 = 11 - 5 \cdot 2$$
Thus, $$-5 \cdot 2 \equiv 1 \pmod{11} \iff 2^{-1} \equiv -5 \pmod{11}$$
What does this mean?  It means that for each $t \in \mathbb{Z}$, 
$$2(-5 + 11t) \equiv 1 \pmod{11}$$
In particular, if $t = 1$, we obtain 
$$2 \cdot 6 \equiv 1 \pmod{11} \iff 2^{-1} \equiv  6 \pmod{11}$$
Why is $2(-5 + 11t) \equiv 1 \pmod{11}$ for $t \in \mathbb{Z}$?  Observe that 
$$2(-5 + 11t) \equiv -10 + 22t \equiv 1 - 11 + 22t \equiv 1 + 11(2t - 1) \equiv 1 \pmod{11}$$
A: It is part of the collection of integer pairs such that:
$$xy\equiv1\mod p$$
in this case with $p=11$.
For $p=11$, they are $1\cdot1,2\cdot6,3\cdot4,5\cdot9,7\cdot8$ and $10\cdot10$
Except for $1$ and $p-1$, each $x$ is paired to a unique $y$.
