If A and B are co-primes, does Ax mod B (where x, any positive int) gives {0,1,2,....,B-1}? If $A$ and $B$ are co-primes (i.e. $\gcd(A,B)=1$), does $A\cdot x \bmod B$ (where $x\in \mathbb N$) give as result an element of $\{0,1,2,....,B-1\}$ ?
Claim: If $A$ and $B$ are coprime, the map
$$\begin{align*}
\{0,\dots,B-1\}\ &amp;\to\ \{0,\dots,B-1\}
,\\
x \ &amp;\mapsto\  A\cdot x\bmod B
\end{align*}$$
is a well-defined bijection.

It is clear directly from the definition of $\bmod$ that $A\cdot x\bmod B$ is indeed between $0$ and $B-1$.
Surjectivity follows, e.g., from Bézout's lemma.
Any surjection between finite sets of the same cardinality is automatically bijective.

In particular, this implies that the map
$$\begin{align*}
\mathbb Z\ &amp;\to\ \{0,\dots,B-1\}
,\\
x \ &amp;\mapsto\  A\cdot x\bmod B
\end{align*}$$
is also surjective.