Given a ring find two divisors of zero $a$ and $b$ such that their sum is a unit. 
Given the ring $\mathbb Z/6\mathbb Z$ find $a$ and $b$, two divisors of zero, such that their sum is a unit.

Given the ring $\mathbb Z/6\mathbb Z = \{0,1,2,3,4,5\}$. So by two divisors of zero what I understand is that $a|6$ and $b|6$, and by their sum is equal to unity $a+b = u$, where $\exists \  v \in \ \mathbb Z/6\mathbb Z $ such that $ uv = vu = 1$.
Pairs of distinct elements $a|6$ and $b|6$, are $1,3$ ; $3,2$ ; $1,2$. Their sum is $1+4 = 5$; $2+3 = 5$; $1+2 = 3$.
$ 5 \cdot 5 = 24 \equiv 1 \mod (6)$. That is, $5$ is the only unit that satisfies the condition.
Did I miss something? Is the approach correct? Do I even understand the exercise? Please feel free to correct me.
Thank you very much. Kind regards.
 A: As pointed out in the comments there seems to be some misconception about the term "zero divisor". To make it perfectly clear: given a ring $R$ an element $a\in R$ is called a zero divisor if there exists $b\in R\setminus\{0\}$ such that $ab=0$. In particular, $0$ is always a zero divisor a rings for which this is the only zero divisor are called integral domains.
Back to $\mathbb Z/6\mathbb Z$. An element $a$ here is a zero divisor iff there is some non-zero $b$ such that $ab=0$. As $\mathbb Z/6\mathbb Z$ is quite a small ring you can simply check all elements to find that $0,2,3,4$ are the only zero divisors (you missed $0$ in your comment; see what I wrote above). Also, $1$ is always a unit and you showed that $5$ is a unit too. So, there are two immediate possiblities: $(a,b)=(2,3)$ or $(a,b)=(3,4)$. As you are only asked to find a pair $(a,b)$ such that $a+b$ is a unit this should suffice.

Two facts to prove which might be a good training for better understanding units and zero divisors:
Fact $\mathbf1$. Let $R$ be a ring. If $u\in R$ is a unit then $u$ is not a zero divisor.
Fact $\mathbf2$. Let $R=\mathbb Z/n\mathbb Z$. Then $R$ decomposes disjointly into its set of units and zero divisors. That means every element of $R$ is either a unit or a zero divisor. (This holds more generally for all finite rings)
