imc

We know that $\gcd (n+57,63)=21$ and $\gcd (n-57,120)=60$ by the Euclidean Algorithm. Hence, let $n+57=21\alpha$ and $n-57=60\gamma$, where $\gcd (\alpha ,3)=1$ and $\gcd (\gamma ,2)=1$. Subtracting the two equations, $38=7\alpha -20\gamma$. Letting $\gamma =2s+1$, we get $58=7\alpha -40s$. Taking modulo $40$, we have $\alpha \equiv 14(\mathrm{mod}40)$. We are given that $n=21\alpha -57>1000$, so $\alpha \ge 51$. Notice that if $\alpha =54$ then the condition $\gcd (\alpha ,3)=1$ is violated. The next possible value of $\alpha =94$ satisfies the given condition, giving us $n=1917$. Alternatively, we could have said $\alpha =40k+14\equiv 0(\mathrm{mod}3)$ for $k\equiv 1(\mathrm{mod}3)$ only, so $k\equiv 0,2(\mathrm{mod}3)$, giving us our answer. Since the problem asks for the sum of the digits of $n$, our answer is $1+9+1+7=\boxed{18}$.