Is there a mathematical way to know the answer? (or you can do it only by trial and error) Could you prove that it is possible or impossible mathematically?
It is possible to arrange all possible topologies and calculate the resistance of each. Nice idea for programming homework.
Proving that something is possible requires only one example. In your case: one resistor between the two poles, all other resistors unconnected (or connected to one pole, etc).
Proving that something is impossible requires an ad-hoc proof or enumerating all possible topologies.
Another possibility would be:
(6//6//6) + 6//(6+6) = 2 + 6//12 = 2 + 4 = 6
BTW, I did note that you're after a mathematical solution, but since I couldn't think of one, I offered this. It would certainly be possible to solve it algorithmically, with iterations, but a single mathematical solution may not be possible? Very interesting question.
This problem is under constrained.. what does 'arranged' mean? Can you use one or four in series-parallel and short the left-over resistors?
It's not possible to have them share power equally, however it is possible to actively use all the resistors. Hint: calculate 1/( 1/9 + 1/18 )
If there is a straightforward mathematical way, I'm not aware of it.
This appears to be related to:
which leads to just twelve graphs for six edges - quite a suprise to me. You will then need to measure n! node pairs.
Oh - I quickly came up with the 'leave 5 unconnected' (a definite cheat) and bridge (not a cheat) circuits. Kudos to the answers where all resistors carry current.