Selection tests for the Balkan Mathematical Olympiad 2013

We have $n=c+7b+49a=a+9b+81c$. This implies that $20(2c-a)=(4a-b)$. Because $0\le a,b,c\le 6$, either $4a-b=2c-a=0$ or $4a-b=20$ and $2c-a=1$.

1. If $4a-b=2c-a=0$, then $b=8c\le 6$ and therefore $a=b=c=0$, that is $n=0$.

2. If $4a-b=20$ and $2c-a=1$, then $a=5$, or $6$. If $a=5$ then $b=0$ and $c=3$ and $n=248$. If $a=6$, then $2c=7$, which is impossible.

Therefore, $n=0$ or $248$.