An integer $x$ gives the same remainder when divided by both $3$ and $6$. It also gives a remainder of $2$ when divided by $4$, can you determine an unique remainder when $x$ is divided by $6$?
I feel like you can't since $x=4q+2$ for integer $q$. Listing out some $x$'s gives $x = 2, 6, 10, 14, 18, \cdots$. When you divide these numbers by 6 you get the remainders $2, 0, 4, 2, 0, 4, \cdots$ and when you divide these numbers by 3, you get $2, 0, 1, 2, 0, 1 \cdots$, so the remainders in common are $2$ and $0$, and so it's not enough to determine an unique remainder.
Could anyone show me a proper argument of this without actually having to list out all the numbers and manually "test" it?