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fshiznit
Jan29-09, 12:57 PM
You receive a box of 9 resistors. All the resistors except one have a value of 1 ohm. With only an ohmmeter, what is the minimum number of measurements needed to identify the odd resistor and find its resistance.

-I've tried going up from one resistor, but i'm not sure what i'm doing past 4 resistors :S..any help is much appreciated!

Gokul43201
Jan30-09, 08:59 AM
-I've tried going up from one resistor, but i'm not sure what i'm doing past 4 resistors :S..any help is much appreciated!What do you mean by "going up from one resistor"? Please be more explicit in your explanation.

fshiznit
Jan30-09, 10:37 AM
so for let's say 3 resistors, you'll need two measurements because say you take an even one and an odd one and get a resistance of 2.1 ohm (that's one measurement) then you take one off to which is the odd one and its resistance (second measurement). but now im thinking, wouldn't the minimum measurements, even for the 9 resistor one, be just two measurements, seeing that you get lucky and pick the odd one when testing the first two resistors? But then i guess the question'd be too easy :S

Redbelly98
Jan31-09, 11:37 AM
As I interpret the question, assuming you are unlucky, what is the minimum number of measurements required?

A naive approach would be to measure them one-by-one and the odd one is the last resistor. However, you have to be a little clever here. Other than measuring one at a time, how else might you approach the problem?

sberxasberxa
Feb2-09, 08:13 PM
would it make sense to put them in groups of 2 in parallel? (thus making 4 groups in series). Would that not only take a max. of 5 measurements depending on your luck?

mplayer
Feb5-09, 12:58 AM
Hint: measure resistances across multiple elements at once and think about what that tells you. You cannot rely on luck to get the minimum number of measurements.

Redbelly98
Feb5-09, 06:57 AM
FYI to everybody:

We really should let fshiznit try to solve the problem, and let him/her respond with questions, before giving further help.