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.
 A: 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 $3^{-1} = 2 \mod 5$.
A: Solving the first two equations simultaneously, you get x = 24(mod 110). Solving this result and the third equation simultaneously, you get x = 134(mod 330).
A: The meaning of modular inverses has been explained. There is also a small error that needs fixing: $\rm\:mod\ 11:\ x\equiv 12\:\Rightarrow\:x\equiv 1\ \ (not\,\ 2).\:$ Also, brute-force Chinese remainder is usually not the most efficient method of solution. For example, here is a simpler method (no inverses needed).
$\rm mod\ 11\!:\ x\equiv 1\:\Rightarrow\: x = 1\!+\!11j.\:$ So $\rm\:mod\ 10\!:\ 4 \equiv x = 1\!+\!11j\equiv 1+j\:\Rightarrow\:j\equiv 3,\:$ so $\rm\:j = 3\! +\! 10k,\:$ thus $\rm\:x = 1\!+\!11j = 1\!+\!11(3\!+\!10k) = 34\!+\!110k.\:$ So $\rm\:mod\ 3\!:\ 2\equiv x\equiv 34\!+\!110k\equiv 1\!-\!k\:\Rightarrow\:k\equiv -1\equiv 2,\:$ i.e. $\rm\:k = 2\!+\!3n.\:$ Hence $\rm\:x = 34\!+\!110k = 34\!+\!110(2\!+\!3n) = 254\!+\!330n$.
