Arrangement Puzzle

[Nameo]

You have 7 resistors that should all have a value of 1 but one has a value greater than one (resistor is a physics term, unimportant though), what is the fewest number of measurements needed to find which one does not have a value of 1 and what it's exacty value is (measuring them independently, grouping four and seeing they are worth 4.2 instead of what should be 4) ?

[gen2]

Answer = 3 measurements are all that are needed.

1) Measure all resistors together in series. That will tell you what the value of the odd resistor is by subtracting 6.

2) Join any 3 in parallel. Join any 2 in parallel. Now join one of the single resistors, the pair and the triplet in series and measure their resistance. If it comes to 1.83333, you know the odd one is the one you left out of this circuit. If not, simple arithmetic should let you work out whether the odd one is the single one in your circuit, one of the pair, or one of the triplet. (1/R = 1/r1 + 1/r2 + 1/r3).

3) If the odd one is in the pair, just measure the resistance of one of them and that will tell you if the measured one or the other one is the odd one out.
If the odd one is in the triplet, let's call them A, B and C, make up a circuit with A and one of the eliminated resistors that we know is 1 ohm in parallel and connect it in series with B (leave out C). You should have a pair and single resistor in series. Measure this combination. If it measures 1.5 you know the odd one is the one you left out of this circuit (C). If not, simple arithmetic should let you work out whether the odd one is the single one in your circuit (B), or one of the pair. (1/R = 1/r1 + 1/r2 ). If one of the pair, you will know which one (A) because you used a known 1 ohm resistor as the other one of them.

[rojash]

That seems an awfully complicated way of doing in. You could do it all connecting the resistors only in series and using a binary chop search to find the odd one. This would also take a max of 3 steps.