SELECTION EXAMINATION

We distinguish three cases mod $3$.

If $n=3k$, then $A=7^{6k}-48(3k)-1=49^{3k}-9\cdot 16k-1$. Since $49\equiv 4(\mathrm{mod}9)$, it follows that $49^3\equiv 4^3(\mathrm{mod}9)\equiv 1(\mathrm{mod}9)$ and hence $9\mid 49^{3k}-1\Rightarrow 9\mid A$.

If $n=3k+1$, then $A=7^{6k+2}-48(3k+1)-1=7^2\cdot 7^{6k}-9\cdot 16k-49=49(7^{6k}-1)-9\cdot 16k$. Since $9\mid 7^{6k}-1$, it follows again that $9\mid A$.

* If $n=3k+2$, then $A=7^{6k+4}-48(3k+2)-1=7^4\cdot 7^{6k}-9\cdot 16k-97$. We have that $7^{6k}\equiv 1(\mathrm{mod}9)\Rightarrow 7^4\cdot 7^{6k}\equiv 7^4(\mathrm{mod}9)\equiv 49^2(\mathrm{mod}9)\equiv 4^2(\mathrm{mod}9)\equiv 7(\mathrm{mod}9)$ and also $97\equiv 7(\mathrm{mod}9)$. Hence $9\mid 7^4\cdot 7^{6k}-97$, that is $9\mid A$.