Find the sum of all $abc$ 
Let $T$ be the set of all triplets $(a,b,c)$ of integers such that $1 \leq a < b < c \leq 6$. For each triplet $(a,b,c)$ in $T$ , take number $a \cdot b \cdot c$. Add all these numbers corresponding to all the triplets in $T$. Prove that the answer is divisible by $7$.

Attempt
Let $S = \displaystyle \sum_{b < c} bc $. Then with no restriction on $a$ we have $M = (1+2+3+4+5+6)S = 21S$. This is the sum of the cases where $a < b, a = b, $ and $a > b$. Now how can I use this to find the case where $a<b$?

For any such triplet $(a, b, c)$ with $a<b<c$, 
we have a corresponding triplet $(7-c, 7-b, 7-a)$ with $7-c<7-b<7-a$.
Since we can pair up all the triplets in this manner, 
and $abc+(7-c)(7-b)(7-a)\equiv 0\pmod{7}$,
the sum of the products of each of the triplets is divisible by 7.
