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Yes.
The first and most obvious point is that any factor of twelve, other than one (leaving 2, 3, 4, 6, 12), is a solution.
Next, realize that the superposition of any two solutions is yet another solution; two arrangements that are balanced individually will be balanced when placed at the same time, provided that it is possible to do so without needing two tubes to occupy the same slot.
For the case 2+3 (to get the five we are looking for in the problem), 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).
So, 1, 2, 5, 8, 9 is one example of a five-slot solution.
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Note that this is the only solution, save rotations of the same pattern.
The obvious generalization that follows is, what numbers of tubes cannot be balanced?
First, we also observe that if x has a solution, 12-x also has one (obtained by swapping tubes and holes). Considering this, the valid factors of 12 (2, 3, 4, 6, 12), and the method for running five tubes that we have just determined, it is obvious that 1 and 11 are the only cases without solutions. In other words, since each number 2-5 has a solution, so do 7-10 (as well as 0, 6, and 12).
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 of equal weight opposite it for balance. |
