I have to find the $M$ and $N$ to be non-free modules over the ring $\mathbb{Z}/6\mathbb{Z}$, such that $M\oplus N$ is free.
My idea was to take $M=2\mathbb{Z}/6\mathbb{Z}$ (That is, the even numbers in the numbers modulo 6), and then take $N=3\mathbb{Z}/6\mathbb{Z}$ (the numbers multiple of three modulo 6).
Then $N$ is not free because the only elements in it are $\{0,3\}$, and none of them is free (if we take $3$, then we could have $r\cdot 3=0$ by letting $r=2$).
Similarly I argue that $M$ is not free. Lastly, it is easy to see that $\mathbb{Z}/6\mathbb{Z}=M\oplus N$ since the only element in common for $N$ and $M$ is zero, and their sum gives the desired result.
Questions; Is the above correct? Which other(easy) examples are there