India — Team Selection Test, 2008

First observe that $A$ and $B$ are separated by $a$ numbers and as such $A \cup B$ is not a set of consecutive integers.

a. ($\Rightarrow$) If $r = t$ and $a = kr$ for some $k \in \mathbb{N}$, then it is clear that $Z$ is covered by the mutually disjoint sets $(A \cup B) + s$, where $s$ ranges over the set
$$\bigcup_{m \in \mathbb{Z}} \{2(k+1)rm, 2(k+1)rm + r, \dots, 2(k+1)rm + kr\} = 2(k+1)rZ \cup \{2(k+1)rZ + r\} \cup \dots \cup \{2(k+1)rZ + kr\}.$$
(Here $xP$ denotes the set $\{x : p \in P\}$.)

b. ($\Rightarrow$) If $r < t$ and $a = k(r + t)$ for some $k \in \mathbb{N}$, then $Z$ is covered by the mutually disjoint sets $(A \cup B) + s'$, where $s'$ ranges the set $\{(r+t)s : s \in Z\} = (r+t)Z$.

a. ($\Leftarrow$) If $r = t$, and $Z$ is covered by mutually disjoint sets of the form $(A \cup B) + s$, for $s$ in a fixed subset of $Z$, then any translate of $A \cup B$ has a gap of size $a+r+1-(r+1)=a$, which should be covered by mutually disjoint sets of size $r$. This proves that $a = kr$, for some $k \in \mathbb{N}$.

b. ($\Leftarrow$) If $r < t$, and $Z$ is covered by a certain class of mutually disjoint translates of $A \cup B$, we will prove that the translates of $A$ and $B$ must alternate in any covering of $Z$ by translates of $A \cup B$. This would imply that $a = k(r + t)$ for some $k \in \mathbb{N}$. If a covering contained two consecutive translates $A + x$ and $A + x + r$ of $A$, it would contain the corresponding matching translates $B + x$ and $B + x + r$ of $B$. But

$B+x = \{a+r+1+x, a+r+2+x, ..., a+r+t+x\}$ and $B+x+r = \{a+2r+1+x, a+2r+2+x, ..., a+2r+t+x\}$. As $a+2r+1+x \le a+r+t+x$, we see that $B+x$ and $B+x+r$ overlap. Hence this is impossible.

Suppose now that a covering of $Z$ contained two consecutive translates $B+x$ and $B+x+t$ of $B$. Then the covering must contain the corresponding matching translates $A+x$ and $A+x+t$ of $A$.

Now
$$A+x = \{1+x, 2+x, ..., r+x\} \text{ and } A+x+t = \{1+x+t, 2+x+t, ..., r+x+t\}.$$
The gap between $A+x$ and $A+x+t$ corresponds to the set $\{r+x+1, r+x+2, ..., t+x\}$ which is of size $t-r$, which is less than $t$. So $A+x$ must be followed immediately by another translate of $A$ itself. But this is exactly what we proved to be impossible just above. Thus two consecutive translates of neither $A$ nor $B$ can occur in any covering of $Z$. This completes the proof.