Let $F: \mathbb{Z} \longrightarrow \mathbb{Z} \times \mathbb{Z}_n$ be the homomorphism of groups given by $$F(a)=(ma,-[a]_n)$$ with fixed $m \in \mathbb{Z}$
What is, up to isomorphism, the quotient group $\mathbb{Z} \times \mathbb{Z}_n/\mathrm{Im}(F)$
Consider the homomorphism $G : \mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/mn\mathbb{Z}$ given by:
$$ (x,y+n\mathbb{Z})\mapsto x+my+mn\mathbb{Z} $$ Left to the reader: Show that this is well defined, that it is a group homomorphism, and that $\mathrm{Im}(F)\subseteq\mathrm{Ker}(G)$.
To show the other inclusion:
Suppose $G(x,y+n\mathbb{Z})=mn\mathbb{Z}$, then $x+my\in mn\mathbb{Z}$ so there is some k such that $x+my=mnk$, so $x=mnk-my=m(nk-y)$, and $y\equiv -(nk-y) \mod n$, that is $$(x,y+n\mathbb{Z})=(m(nk-y),-(nk-y)+n\mathbb{Z})=F(nk-y).$$
Now we have $\mathrm{Im}(F)=\mathrm{Ker}(G)$, so by the first isomorphism theorem we get
$$(\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z})/\mathrm{Im}(F)\cong \mathrm{Im}(G)$$
Now $\mathrm{Im}(G)=\mathbb{Z}/mn\mathbb{Z}$ because for all $x\in\mathbb{Z}$, $G(x,n\mathbb{Z})=x+mn\mathbb{Z}$
Note:
To get to the homomorphism above, I checked what $(x_1,y_1+n\mathbb{Z})+\mathrm{Im}(F)=(x_2,y_2+n\mathbb{Z})+\mathrm{Im}(F)$ in the quotient group means, which turned out to be (after some working out)
$$ \left\{\begin{aligned} &x_1\equiv x_2&&\mod m\\ &x_1+my_1\equiv x_2+my_2&&\mod mn \end{aligned}\right. $$
And it turned out that the first congruence follows from the second, and the second congruence gives the homomorphism defined above.