Consider the two sets
$$A_r = \{x \in \mathbb{Z} \enspace | \enspace x \enspace = \enspace 5q + r, \enspace 0 \enspace \leq \enspace r \enspace < \enspace 5\}$$
$$\Lambda = \{A_r\}$$
We must show that $\Lambda$ is a partition of the set $\mathbb{Z}$.
I know that I must show that $\emptyset \notin \Lambda$, that two sets $A_r \cap A_k = \emptyset $ where $ 0 \leq k,r < 5$, which can be done by assuming that if $x \in A_r,A_k$, then $A_r = A_k$ where $k$ is an arbitrary integer; and, finally, the union of all the sets in $\Lambda$ is precisely the set $\mathbb{Z}$.
I am not allowed to use the division algorithm, which would make quick work of the first property of a partition. I assume that one plan of attack to show that the first property is fulfilled is by exhaustion; that is, for each $r$, we can find an integer $q$ where $x \in A_r$. Would this be right? Is there an alternative route?
As for the other two properties, I am not sure how to start. Any insights?
There's an alternative route, kind of, although it involves sneakily proving the division algorithm.
Everything in any of those sets is an integer. That's just obvious.
The sets do cover $\mathbb{Z}$. Indeed, suppose $n$ were the least nonnegative integer not in some $A_r$. Note that $n \geq 5$ since $n \in A_n$ for $n \in \{0, 1, 2, 3, 4 \}$.
Now, $n-5$ is in some $A_r$ - say it's $5q+r$ - so $(n-5)+5$ is in the same $A_r$ (simply express it as $5(q+1)+r$). This is a contradiction, so all positive integers are in some $A_r$. Similar reasoning for the negatives.
The sets are all distinct. Indeed, suppose $x \in A_r$ and $x \in A_s$. Then $x = 5p+r = 5q+s$, so $5(p-q) = s-r$. But $r, s \in \{0, 1, 2, 3, 4 \}$, and the only way $s-r$ can be divisible by 5 if $r, s$ come from that list is if $s-r=0$.