How many arrangements for 7 seven numbers so that every four consecutive have sum divisible by 3?

Question

In how many ways can one arrange the numbers 100, 110, 120, 130, 140, 150, and 160 such that the sum of every four consecutive numbers is divisible by 3?

Answer 1

With the rule as stated there are no arrangements that work.

Proof: For any five integers, $a,b,c,d,e$, such that $3\,|\,(a+b+c+d)$ and $3\,|\,(b+c+d+e)$ we must have $3\,|\,(e-a)$. Thus, if your supposed arrangement is $n_1, n_2, \cdots, n_7$ we must have $$n_1\equiv n_5\quad n_2\equiv n_6\quad n_3\equiv n_7\pmod 3$$

Now, $\pmod 3$ your list is $$\{1,2,0,1,2,0,1\}$$

So we note that there are three $1's$ and two each of $0's$ and $2's$.

The block $n_1,n_2,n_3$ repeats as $n_5,n_6,n_7$ $\pmod 3$ with $n_4$ being the odd man odd. Since $1$ is only residue that appears an odd number of numbers, we must have $n_4\equiv 1 \pmod 3$. But then $n_1,n_2,n_3$ are $0,1,2$ in some order which makes $$n_1+n_2+n_3+n_4\equiv 0+1+2+1\equiv 1 \not \equiv 0 \pmod 3$$ and we are done.