Let $F = \lbrace S_1, S_2, \dots, S_n \rbrace$, where $S_i \subset \lbrace 1, 2, \dots, 3m\rbrace$ and define a function $f: F \to \mathbb{N}$ by $$ f(S_i) = \sum_{j \in S_i} (n+1)^{3m-j} $$
then $S_i \sqcup S_j = S \iff f(S_1) + f(S_j) = f(S)$, where $\sqcup$ is disjoint union.
I've already proven that $S_i \sqcup S_j = S \implies f(S_1) + f(S_j) = f(S)$, but I'm not sure how to prove the other direction.
This is not homework. I'm doing preparations for an oral exam and a proof that relies on this was skipped by the course notes.
Let $i \neq j$ and define $T=S_i \cap S_j$. Then we can write $$f(S_i)=\sum_{k \in S_i-T}(n+1)^{3m-k}+\sum_{k \in T}(n+1)^{3m-k}.$$ Likewise $$f(S_j)=\sum_{l \in S_j-T}(n+1)^{3m-l}+\sum_{l \in T}(n+1)^{3m-l}.$$ Then $$f(S_i)+f(S_j)=\sum_{k \in S_i-T}(n+1)^{3m-k}+\sum_{l \in S_j-T}(n+1)^{3m-l}+2\sum_{c \in T}(n+1)^{3m-c}.$$ Let $S=S_i \cup S_j$, then $S=(S_i-T)\cup (S_j-T) \cup T$. Therefore $$f(S)=\sum_{k \in S_i-T}(n+1)^{3m-k}+\sum_{l \in S_j-T}(n+1)^{3m-l}+\sum_{c \in T}(n+1)^{3m-c}.$$ So for $f(S)=f(S_i)+f(S_j)$, we need to have $$\sum_{c \in T}(n+1)^{3m-c}=0.$$ Since we are summing positive integers therefore the only way this sum could be zero is when $T=\phi$, in which case $S_i \cap S_j=\phi$ and $S$ would be the disjoint union of $S_i$ and $S_j$.