= Solution
Use the <Chinese remainder theorem> to separate relatively prime factors:
$$
\mathbb Z/60\mathbb Z\cong\mathbb Z/4\mathbb Z\oplus\mathbb Z/3\mathbb Z\oplus\mathbb Z/5\mathbb Z,
\qquad
\mathbb Z/21\mathbb Z\cong\mathbb Z/3\mathbb Z\oplus\mathbb Z/7\mathbb Z.
$$
These five cyclic prime-power summands are the elementary-divisor form. Leave one $\mathbb Z/3\mathbb Z$ summand aside and combine the factors of orders $4,3,5,7$, which are pairwise coprime. Their <direct sum of modules> is cyclic of order $420$. The invariant-factor canonical form is therefore
$$
\boxed{A\cong\mathbb Z/3\mathbb Z\oplus\mathbb Z/420\mathbb Z,\qquad3\mid420.}
$$
Its <group order> is $3\cdot420=1260=60\cdot21$, and its <exponent of a finite group> is $420=\operatorname{lcm}(60,21)$, providing independent checks on the decomposition. Since its exponent is smaller than its order, $A$ itself is not cyclic.
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