There is a question in my book ( Pathfinder for Olympiad mathematics):
Let T be the set of all triplets (a, b, c) of integers such that
$1 ≤ a ≤ b ≤ c ≤ 6$. For each triplet (a, b, c) in T, take the number $a × b × c$ and add all
these numbers corresponding to all the triplets in T. Prove that this sum is divisible by 7.
I tried but couldn't get through the question. The solution says this :
If (a, b, c) is a valid triplet then $(7 – c, 7 – b, 7 – a)$ is also a valid triplet as $1 ≤ (7 – c) ≤ (7 – b) ≤ (7 – a) ≤ 6 \;
And \; (7 – b) ≠ b$, etc.
Let $S = \sum_{1 ≤ a ≤ b ≤ c ≤ 6} (abc)$ ,
then by the above $S= \sum_{1 ≤ a ≤ b ≤ c ≤ 6}(7-a)(7-b)(7-c)$.
And then the above two equations were added and we got the desired answer.
But I would like to know that is there any alternate method of solving this question ( please don't tell me to multiply those digits and get the numbers and add them and check the divisibility) that can be easily understood by a high school student ?
Thanks in advance.
Best Answer
Here is another answer that uses the observation:
(A) $(7-a)(7-b)(7-c) \equiv -abc \mod 7$.
But also note (B): $$\sum_{a,b,c} abc = \sum_{a,b,c} (7-c)(7-b)(7-a)$$
(because the sets $\{(a,b,c)$; $1 \le a\le b\le c\le 6\}$ and $\{(7-c, 7-b, 7-a)$; $1 \le a \le b \le c \le 6\}$ are clearly the same)
But putting (A) and (B) together gives $\sum_{a,b,c} abc =$ $\sum_{a,b,c} (7-c)(7-b)(7-a) \equiv_7 -1 \left(\sum_{a,b,c} abc\right)$. Thus we conclude from this string:
$$\sum_{a,b,c} abc \equiv_7 -1 \sum_{a,b,c} abc$$
and so $\sum_{a,b,c} abc$ has to be 0 mod 7.