Estonian Mathematical Olympiad

Taking $n=10$ gives $2^n+n^3=1024+1000=2024=88\cdot 23$, so $88$ divides $2^{10}+10^3$.

We consider divisibility by $11$ and by $8$ separately. By Fermat's little theorem, we have $2^{10}\equiv 1(\mathrm{mod}11)$, whereas $10^3\equiv (-1{)}^3=-1(\mathrm{mod}11)$. In summary $2^{10}+10^3\equiv 1-1=0(\mathrm{mod}11)$, meaning that $11$ divides $2^{10}+10^3$. As $8$ divides both $2^{10}$ and $10^3$, it divides the sum $2^{10}+10^3$. Thus $88$ divides $2^{10}+10^3$, as desired.