I'm reading a book and the author leaves as an exercize the proof of the following, assuming that $n\ge 1$ is an integer and $b\in B,c\in C$
If $A=B\oplus C$, then $n\mid (b+c)$ iff $n\mid b$, $n\mid c$.
The implication $\Leftarrow$ is clear to me, while I don't see how to prove the converse.
Recall that, if $G$ is an abelian group, $n\mid x$ means that there exists $y\in G$ s.t. $x=ny$.
Side note: It would probably be a good idea to say what $n,b,c$ are. I'm assuming $n\in \Bbb N$, $b\in B$ and $c\in C$, with $b+c$ being the element $(b,c)\in B\oplus C$. Or something similar. Maybe $b\in B\times \{0\}$ and similarly for $c$ makes more sense, but the result is the same.
If $n\mid (b+c)$, then there is an $(x,y)\in B\oplus C$ such that $n(x,y)=(b,c)$. However, $n(x,y)=(nx,ny)$, showing that $nx=b, ny=c$.