Inverse bit in Chinese Remainder Theorem I need to solve the system of equations:
$$x \equiv 13 \mod 11$$
$$3x \equiv 12 \mod 10$$
$$2x \equiv 10 \mod 6.$$
So I have reduced this to 
$$x \equiv 2 \mod 11$$
$$x \equiv 4 \mod 10$$
$$x \equiv 2 \mod 3$$
so now I can use CRT. So to do that, I have done
$$x \equiv \{ 2 \times (30^{-1} \mod 11) \times 30 + 4 \times (33^{-1} \mod 10) \times 33 + 2 \times (110^{-1} \mod 3) \times 110 \} \mod 330$$
$$= \{ 2 (8^{-1} \mod 11) \cdot 30 + 4(3^{-1} \mod 10)\cdot33 + 2(2^{-1} \mod 3) \cdot 110 \} \mod 330$$
but now I'm stuck on what to do. What do the inverse bits means? If I knew that I could probably simplify the rest myself.

Consider $3 \mod 5$. If I multiply this by $2$, I get $2 \cdot 3 \mod 5 \equiv 1 \mod 5$. Thus when I multiply by $2$, I get the multiplicative identity. This means that I might call $3^{-1} = 2 \mod 5$.
