= Solution
The <Fundamental theorem of finitely generated abelian groups> states that every such <group> is isomorphic to
$$
\mathbb Z^r\oplus\mathbb Z/d_1\mathbb Z\oplus\cdots\oplus\mathbb Z/d_s\mathbb Z,
\qquad 1<d_1\mid d_2\mid\cdots\mid d_s,
$$
with uniquely determined rank and invariant factors, allowing an empty finite part. Modulo $pA$, the free summand contributes $(\mathbb Z/p\mathbb Z)^r$. If $A/pA=0$ for some prime, this forces $r=0$, so $A$ is finite. Conversely, if $A$ is finite, choose a prime $p$ coprime to $|A|$. Bezout's identity gives an integer $b$ with $bp\equiv1\pmod{|A|}$. Since $|A|$ kills every element, multiplication by $p$ is onto, with inverse multiplication by $b$. Hence $pA=A$. This proves \b[$\boxed{A\text{ finite}\iff A/pA=0\text{ for some prime }p}$] under finite generation.
For the requested nonzero example use the additive <divisible group> $A=\mathbb Q$. Each rational $q$ equals $p(q/p)$, so $p\mathbb Q=\mathbb Q$ for every prime. It is not finitely generated: a finite list of rational generators has a common denominator $D$, and every integral combination then lies in $D^{-1}\mathbb Z$. The rational $1/(2D)$ does not lie there. This proves non-finite-generation directly and shows why the hypothesis in <finite generation detected by a prime quotient> is essential.
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