Sufficient conditions such that $A+xB= \mathbb{Z}/n\mathbb{Z} $ I'm working in an exercise and I need some results about additive theory of numbers, I encountered this problem:
Given an element $x\in \mathbb{Z}/n\mathbb{Z}$ and two subsets $A$ and $B$ of $ \mathbb{Z}/n\mathbb{Z} $ can someone help me find relatvely strong sufficient conditions about $A$ and $B$ so that every element of $ \mathbb{Z}/n\mathbb{Z} $ can be written as $a+bx$ with $a\in  A$ and $b\in B$.
The numbers $n$ and $x$ are fixed,
Any help will be highly appreciated,thanks.
 A: The first step would be the case $x | n$. I really hope i didn't mess up anything:
Then we can assume $B \subset \{1,2, \dotsc, \frac{n}{x}\}$, because $\frac{n}{x}x=0$. Let $g$ be the greatest gap (with no gap at all being gap 1) within the numbers in $B$. Then it is sufficient and necessary that $A$ contains $1,2, \dotsc, \lfloor \frac{gx}{2} \rfloor$ modulo $gx$.
Maybe we should consider some example what the greatest gap is.
If $\frac{n}{x}$ is $5$ and $B=\{1,3\}$ then the greatest gap is $3$, because we have 2 consecutive numbers missing: $4,5$
Indeed let $n=15, x=3, B = \{1,3\}$. This gives us $g=3$ and $\lfloor \frac{gx}{2} \rfloor=4$. We have $Bx=\{3,9\}$ and we see that with $A=\{1,2,3,4\}$ we have $A+\{3,9\} = \mathbb Z/15\mathbb Z$.
With $A = \{1,2,3\}$ we don't succeed: $A+\{3,9\}$ does not contain $13$ and $14$.
If $x$ does not divide $n$, we could use $d = gcd(x,n)$. But we encounter the following problems: The sequence $x \mod n,2x \mod n,3x \mod n, \dotsc$ is not increasing anymore, so the gaps within the numbers of $B$ are not directly related to gaps within the numbers of $Bx$.
