= Solution
In the intended range $0\le k\le n$, division by the positive <binomial coefficient> $\binom nk$ gives
$$
\frac{n-2k-1}{k+1}=1,
\qquad n=3k+2.
$$
Consequently the intended <remainder> is
$$
\boxed{n\bmod3=2.}
$$
There 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 $g(k)=\binom nk=0$, regardless of $n\bmod3$. For example, $n=3,k=4$ satisfies the literal equality but has $n\bmod3=0$. Thus the literal extended-domain premise alone does not determine a <remainder>; the boxed conclusion requires $0\le k\le n$, or equivalently here $\binom nk\ne0$.
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