Such colours

We categorise the numbers mod 3:

$\begin{aligned} 21,51,81 &\equiv 0 (\text{mod} 3) \\ 31,61 &\equiv 1 (\text{mod} 3) \\ 41,71 &\equiv 2 (\text{mod} 3) \end{aligned}$

Call an arrangement of the seven numbers $a_1,a_2,\ldots,a_7$ . We have:

$\begin{aligned} a_1+a_2+a_3+a_4& &\equiv 0 (\text{mod} 3) \\ a_2+a_3+a_4& +a_5 &\equiv 0 (\text{mod} 3) \end{aligned}$

hence $a_1\equiv a_5 (\text{mod} 3)$ . Likewise, $a_2\equiv a_6 (\text{mod} 3)$ and $a_3\equiv a_7 (\text{mod} 3)$ . Thus $a_4$ , the one remaining number, must be in the class $0 (\text{mod} 3)$ .

The numbers $a_1,a_2,a_3$ must each be put in a different residue class, which we can pick in $3!$ ways. Each residue class can then be arranged at will, giving $3!\times 2!\times 2!$ possibilities. Hence in total we have $3!3!2!2!=\boxed{144}$ arrangements.

These numbers leave when divided by 3. We have 3 0s (21, 51, 81), 2 1s (31, 61), and 2 -1s (41, 71).

Any combination of the following form will do: 1 -1 0 0 1 -1 0 or -1 1 0 0 -1 1 0 or 0 1 -1 0 0 1 -1 or 0 -1 1 0 0 -1 1

As we can observe it is needed a cycle of 1 -1 0 0 or -1 1 0 0.

For each of the 4 possibilities listed, the permutations are 3! * 2! * 2!=6 * 2 * 2=24 (3! for the 3 0s, and 2! for both 1s and -1s)

There are two more possibilities: 1 0 -1 0 1 0 -1 and -1 0 1 0 -1 0 1

Finally, the total number of ways is 6 * 24 = 144

Scientific I would tell.

From 21, 31, 41, 51, 81, 61, 71 of 144, 204, 234, 264 coded 1234756

To 81, 71, 61, 51, 21, 41, 31 of 264, 204, 174, 144 coded 7654132,

There are a total of 144 out of 5040 which satisfied.

Answer: 144