Is there any quick way to classify a group from list of abelian groups? I'm trying to solve the following problem

Classify group of order $720$ which has exactly $14$ elements of order $6$

By using fundamental theorem of abelian groups, I found following $10$ abelian groups of order $720$.

*

*$\mathbb{Z}_{16}\times\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3$.

*$\mathbb{Z}_{16}\times\mathbb{Z}_5\times\mathbb{Z}_9$.

*$\mathbb{Z}_4\times\mathbb{Z}_4\times\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3$.

*$\mathbb{Z}_4\times\mathbb{Z}_4\times\mathbb{Z}_5\times\mathbb{Z}_9$.

*$\mathbb{Z}_2\times\mathbb{Z}_8\times\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3$.

*$\mathbb{Z}_2\times\mathbb{Z}_8\times\mathbb{Z}_5\times\mathbb{Z}_9$.

*$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4 \times\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3$.

*$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4 \times\mathbb{Z}_5\times\mathbb{Z}_9$.

*$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2 \times\mathbb{Z}_5\times\mathbb{Z}_3\times\mathbb{Z}_3$.

*$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2 \times\mathbb{Z}_5\times\mathbb{Z}_9$.

How to classify the one asked in question? What is the smartest way to do this? Do we have to count the number of elements in each group individually?
 A: Since each of your direct factors has prime power order, an element has order dividing $6$ if and only if its coordinates have order $1$, $2$, or $3$. It will have order exactly $6$ if at least one coordinate has order $2$, and at least one has order $3$. Equivalently, the projections onto the $2$-part and the $3$-part must be elements of order $2$ and $3$, respectively; the projection onto the $5$-part is of course trivial. So the number of elements of order $6$ is the product of the number of elements of order $2$ in the $2$-part, and the number of elements of order $3$ in the $3$-part.
So if your $3$-part is isomorphic to $\mathbb{Z}_9$, you have exactly two choices for that part (the two elements of order $3$), whereas if your $3$-part is isomorphic to $\mathbb{Z}_3\times\mathbb{Z}_3$ you have $8$ possibilities for that part (every element except the identity).
Similarly with the $2$-part: $\mathbb{Z}_{16}$ gives exactly one element of order $2$;  for $\mathbb{Z}_4\times\mathbb{Z}_4$ you have three; for $\mathbb{Z}_2\times\mathbb{Z}_8$ you likewise have three; for $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4$ you have seven; and for $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ you have  fifteen.
Clearly we can discard those final two groups in your list: too many elements of order $6$. If the $3$-part is $\mathbb{Z}_3\times\mathbb{Z}_3$ then the number of elements of order $3$ is $8$, so the number of elements of order $6$ is a multiple of $8$, and so cannot be $14$.
So the $3$-part must be cyclic, and there have to be exactly seven elements of order $2$ in the $2$-part. That means three direct factors in the $2$-part.
