Greece — 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 $49 \equiv 4 \pmod{9}$, it follows that $49^3 \equiv 4^3 \pmod{9} \equiv 1 \pmod{9}$ 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 $7^{6k} \equiv 1 \pmod{9} \Rightarrow 7^4 \cdot 7^{6k} \equiv 7^4 \pmod{9} \equiv 49^2 \pmod{9} \equiv 4^2 \pmod{9} \equiv 7 \pmod{9}$ and also $97 \equiv 7 \pmod{9}$.
Hence $9 \mid 7^4 \cdot 7^{6k} - 97$, that is $9 \mid A$.