Question: Let $k \geq 2$ and $r \geq 4$ be two natural numbers. We are given eight integers $\nu_{ij} \geq 0$ for every $1 \leq i \leq k$ and $1 \leq j \leq r$ such that the following two conditions holds:
For every $r$-tuple of integers $(d_{1}, d_{2}, \ldots, d_{r})$, there exists disoint subsets $B, C$ of $\{1, 2, \ldots, r\}$ with $|B| \not\equiv |C| \hspace{1mm} (\text{mod } 3)$ such that $$ \sum_{j\in B} \nu_{ij} d_{j} \equiv \sum_{j \in C} \nu_{ij} d_{j} \hspace{1mm} (\text{mod } 9). $$
For every $1 \leq j \leq r$, there exists $1 \leq i \leq k$ such that $\nu_{ij} \not\equiv 0 \hspace{1mm} (\text{mod }9)$
We want to show that all the $\nu_{ij} \equiv 0 \hspace{1mm} (\text{mod } 3)$ for every $1 \leq i \leq k$ and $1 \leq j \leq r$.
My Thoughts
A. I see why in condition 1 above, we would enforce that $|B|$ and $|C|$ need not be equal modulo three, because otherwise solutions in the case of the tuples when $c_{1} = c_{2} = \ldots = c_{r}$ may be trivially found. For instance, when $r = 4$ and $c_{1} = c_{2} = c_{3} = c_{4}$, we may just take $B$ and $C$ to have two members each.
B. Negation of 2 is just the case when the hypothesis in the question is true but the conclusion in the question is not true - which is being eliminated.
For instance, if there existed a $1 \leq j_{0} \leq r$ such that for every $1 \leq i \leq k$, we have $\nu_{ij} \equiv 0 \hspace{1mm} (\text{mod } 9)$, then regardless of the tuple $(c_{1}, c_{2}, \ldots, c_{r})$ given to us, we keep choose $B = \{j_{0}\}$, $C = \emptyset$ and since $1 \not \equiv 0 \hspace{1mm} (\text{mod } 3)$, we will easily have hypothesis of the question.
C. My only other intuition is that because there are so many equations, one for each tuple $(c_{1}, c_{2}, \ldots, c_{r})$, each of the $\nu_{ij}$ reduced to being trivial modulo $3$.
I would appreciate any thoughts on this.
