Solution 3077c2cd-aaa6-4675-ab80-f37bf6439897

Transitivity Property of Divisibility

Since $a\mid b$, by the definition of divisibility there exists an integer $m$ such that
\[
b=am.
\]
Similarly, since $b\mid c$, there exists an integer $n$ such that
\[
c=bn.
\]
Substituting $b=am$ into the second equation gives
\[
c=(am)n=a(mn).
\]
Because $m,n\in\mathbb Z$, their product $mn$ is also an integer. Thus $c$ is an integer multiple of $a$, so
\[
\boxed{a\mid c}.
\]