I have to solve $x$ given the next three equations:$x=2\mod 7$, $x=3\mod 11$ and $x=4\mod 13$.
What I tried: first try to find $x$ that satisfies the first two equations. There are $a,b\in \mathbb Z$ such that $7a+11b=1$. So $x=21a+22b$. Then I thought that if I add a multiple of $77$ to $x$, then $x\mod 77$ won't change. But here I got stuck.