The superposition of any two solutions is yet another solution, so given that the factors > 1 of 12 (2, 3, 4, 6, 12) are all solutions, the only thing to check about, for example, the proposed solution 2+3 is that not all ways of combining 2 & 3 would have centrifuge tubes from one subsolution occupying the slot for one of the tubes in another solution. For the case 2+3, there is no problem: Place 3 tubes, one in every 4th position, then place the 4th and 5th diametrically opposed (each will end up in a slot adjacent to one of the first 3 tubes). The obvious generalization is, what are the numbers of tubes that cannot be balanced? Observing that there are solutions for 2,3,4,5,6 tubes and that if X has a solution, 12-X has also one (obtained by swapping tubes and holes), it is obvious that 1 and 11 are the only cases without solutions.

Here is how this problem is often solved in practice: A dummy tube is added to produce a total number of tubes that is easy to balance. For example, if you had to centrifuge just one sample, you'd add a second tube opposite it for balance.

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