Can I always say that any "custom" operation in $Z_n$ is commutative and associative? I have a lot of exercises that say something similiar: 

Given, in the set $Z_{15}$ the following binary operation $*$ $$\forall a,b \in Z_{15}, a*b = \overline6(a+b)\ -\ \overline5ab$$

This is just an example, the operation could be any other operation that does additions and multiplications (without involving inverses). Can I always say (write down in exercise) without manually proving that it is associative and commutative  because $Z_{15}$ is a commutative and unary ring?
 A: Something you can generally do, is looking for a bijection with a set having a binary operation $\oplus$ of which you know that it has the desired properties. Here it is rather evident: Let
$$
f\colon \mathbf Z/15\rightarrow \mathbf Z/3\times\mathbf Z/5
$$
be the map that sends $\overline a$ to the pair $(a\bmod 3,a\bmod 5)$. This is well-defined as $3$ and $5$ divide $15$. The Chinese remainder theorem states that $f$ is a bijection, but it can be easily seen: if $a\bmod3=0$ and $a\bmod5=0$ then $a$ is divisible by $3$ and by $5$. Since $3$ and $5$ are relatively prime, $a$ is also divisible by $15$, i.e., $\overline a=0$. This proves that the kernel of the group morphism $f$ is zero. Hence $f$ is injective. Since the domain and codomain of $f$ are finite and of same cardinality, $f$ is a bijection.
Now, since $6\equiv 0\bmod3$, $-5\equiv1\bmod3$, $6\equiv1\bmod5$ and $-5\equiv 0\bmod5$, one has
$$
f(\overline a*\overline b)=f(\overline 6(\overline a+\overline b)-\overline 5\overline a\cdot \overline b)=
(0(\overline a+\overline b)+\overline a\cdot\overline b,\overline a+\overline b-0\overline a\cdot\overline b)=
(\overline a\cdot\overline b,\overline a+\overline b),
$$
writing $\overline{\cdot}$ for reductions modulo $15$, $3$, and $5$ in order to simplify notation.
One notices that
$$
f(x*y)=f(x)\oplus f(y)
$$
where $\oplus$ is the binary operation on $\mathbf Z/3\times\mathbf Z/5$ defined by
$$
(x_1,x_2)\oplus(y_1,y_2)=(x_1y_1,x_2+y_2).
$$
Since $f$ is a bijection, all properties of $\oplus$ are also satisfied by $*$! For example, since one knows that $\oplus$ is associative, $*$ is associative, and so on.
In fact, this his how teachers construct unusual binary operations for you to check commutativity, associativity or other properties!
