I am trying to count the number of solutions to the following congruence relation. $$x_1^3+x_2^3+x_3^3+\cdots+x_k^3 \equiv 0 \bmod 3^{\lambda}$$ where $0\leq x_i<3^{\lambda}~~\forall ~i \in [1,k]$
Let $C(\lambda,k)$ denote the number of solutions to the above congruence. I wrote a code to generate some results. It appears that $C(1,k)=3^{k-1}$. $C(\lambda,k)$ seems to be always divisible by $3^{(\lambda-1)(k-1)}$. Is there possible to explicitly find $C(\lambda,k)$ for any $\lambda$ and $k$ without exhaustive search?
Thanks for your help.