Is $\mathbb{Z}^n/(m,a_2,...,a_2)\mathbb{Z}\cong\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}^{n-1}$ for $0<m<a_2,...,a_n\in\mathbb{Z}$ (as $\mathbb{Z}$-modules)? If not, how can you compute something like $\mathbb{Z}^3/(2,0,3)\mathbb{Z}$? For example the $\mathbb{Z}$-linear map $\mathbb{Z}^3\longrightarrow \{0\}\oplus\mathbb{Z}^2, (a,b,c)\mapsto (a,b-2a,c-3a)$ induces an isomorhphism $\mathbb{Z}^3/(1,2,3)\mathbb{Z}\cong\mathbb{Z}^2$, because for
\begin{equation} M= \begin{pmatrix} 0 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \end{pmatrix} \ \end{equation} holds $\operatorname{rank}(M)=2$ and $\ker(M)=(1,2,3)\mathbb{Z}$. But for $\mathbb{Z}^3/(2,0,3)\mathbb{Z}$ or other examples I don't know enough module theory.
In fact $\Bbb Z^3/\Bbb Z(a,b,c)$ where $(a,b,c)$ is a nonzero vector, is isomorphic to $\Bbb Z^2\oplus \Bbb Z/g\Bbb Z$ where $g=\gcd(a,b,c)$. This is a special case of the Smith normal form algorithm for calculating the structure of quotients of $\Bbb Z^n$.
Here $g=1$ and $\Bbb Z^3/\Bbb Z(2,0,3)\cong\Bbb Z^2$, and free generators of this group are given by $\overline{(1,0,1)}$ and $\overline{(0,1,0)}$. To see this, note that the vectors $(2,0,3)$, $(1,0,1)$ and $(0,1,0)$ are free generators of $\Bbb Z^3$.