 Solution

In the intended range $0\le k\le n$, division by the positive binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ givesConsequently the intended remainder isThere is a necessary qualification to the printed unrestricted domain. If binomial coefficients outside $0\le k\le n$ are extended by zero, then every $k>n$ satisfies $g(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$, regardless of $n\mathrm{mod}3$. For example, $n=3,k=4$ satisfies the literal equality but has $n\mathrm{mod}3=0$. Thus the literal extended-domain premise alone does not determine a remainder; the boxed conclusion requires $0\le k\le n$, or equivalently here $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\ne 0$.