The Case Of The Spinning Spaceship (Continued)

In my previous solution to The Case Of The Spinning Spaceship , I explained Archimede's Law Of The Lever :

The turning effect , or moment , on a lever, is equal to the force multiplied by the perpendicular distance from the fulcrum to the point of application . For example, a mass placed on the end of a lever a distance $d$ from the fulcrum is balanced by half the mass applied at a point $2d$ from the fulcrum, on the opposite arm. (The acceleration component of the force is provided by gravity.)

We then calculated the moment for the spaceship:

$m_a=4d\vec a-4d\vec a=0$

$m_b=0+d_{engine}\vec b-2d_{engine}\vec b+3d_{engine}\vec b-4d_{engine}\vec b+5d_{engine}\vec b-6d_{engine}\vec b+7d_{engine}\vec b=4d_{engine}\vec b$

In the second equation, $d_{engine}$ is equal to the distance between consecutive engines, and $nd_{engine}$ is the distance between the $n$ th oar and the stern, which has position $d_{engine}=0$ .

We must now solve for when the moment is $0$ . As the distance $d$ remains unchanged, the first equation is still equal to $0$ . This leaves us with the second equation, which, when the left-right position of the engines is unknown and the moment is assumed to be $0$ , takes the form:

$m_b=0=0\pm d_{engine}\vec b\pm2d_{engine}\vec b\pm3d_{engine}\vec b\pm4d_{engine}\vec b\pm5d_{engine}\vec b\pm6d_{engine}\vec b\pm7d_{engine}\vec b$

Dividing both sides by $d\vec b$ gives:

$0\pm1\pm2\pm3\pm4\pm5\pm6\pm7=0$

Now, the sum of $[0, 1, 2, 3, 4, 5, 6, 7]$ is $28$ , which means that the sum of the moments of one side of engines must be $14$ . Thus, this question is equivalent to separating the numbers $[0, 1, 2, 3, 4, 5, 6, 7]$ into two sets with sum $14$ .

Begin with the set containing 0. If it also contains $1$ , then the other two numbers must be $[6, 7]$ . If it contains $2$ but not $1$ , then the other two numbers are either $[4, 6]$ or $[5, 7]$ . If it contains $3$ but not $2$ and/or $1$ , then the other two numbers must be $[5, 6]$ . By similar reasoning, we cannot find a similar set for $4, 5, 6, 7, 8$ . Thus, the $\boxed{4}$ sets $[0, 1, 6, 7; 2, 3, 4, 5]$ , $[0, 2, 4, 6; 1, 3, 5, 7]$ , $[0, 2, 5, 7; 1, 3, 4, 6]$ , and $[0, 3, 5, 6; 1, 2, 4, 7]$ are the only possible arrangements.