12 rubies weighing problem/riddle

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To identify the special ruby among 12 identical rubies using a balance scale only three times, the merchant can follow a strategic weighing process. First, divide the rubies into three groups of four. Weigh two of the groups against each other. If they balance, the special ruby is in the group not weighed. If they don’t balance, the heavier or lighter side indicates the group containing the special ruby.Next, take the group identified as containing the special ruby and divide it into three groups of one, one, and two rubies. Weigh the two single rubies against each other. If they balance, the special ruby is one of the two remaining. If they don’t, the heavier or lighter side indicates the special ruby.Finally, if the special ruby is among the two, weigh one against the other to determine which is the special ruby and whether it is heavier or lighter. This method efficiently narrows down the possibilities within the constraints of three weighings.
hussein236
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i searched for this one if it is posted before, apparently not, but if it is, sorry for that.

a merchant has 12 identical rubies with equal masses that he wants to put on display, however, he knows that one of these rubies is special in the sense that its either heavier or lighter than the rest. he wants to know which one of them is the special one AND if it is heavier or lighter. however, he has a balance scale that he can use only 3 times. how would he identify the special ruby and know if its heavier or lighter.

notice:
the scale can only tell which side is heavier, the merchant can't measure the angle at which its sloped.
the rubies are of identical shapes, but they can be labeled.
the merchant doesn't have any other tools but the balance scale.
 
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Fun stuff, this is.

Merchant weighs six rubies against six rubies. (2 weigh-ins remaining)
He takes the heavier six and divides them, three against three. He weighs these against each other. (1 weigh-in remaining)
The heavier three are split again. This time just one to one. (no weigh-ins remaining)
If the balance is not equal, the heavier one is obvious. If the balance is equal, then, by elimination, the last, unweighted ruby is the heaviest.

However, this only works if you know that the odd ruby is also the heaviest. I was just thinking about it and if, in the previous step, the balance is not equal and you don't know if the ruby is supposed to be heavier or lighter then you can't determine which the odd ruby is. I'd be very interested in hearing the solution if this is the case.
 
Last edited:
No

first move: 1234 x 5678

If 1234 lighter:

Second move 125 x 369

if 125 lighter

Third move 1 x 2

If 1 lighter: solution 1 lighter

If 1 heavier: solution 2 lighter

if equal: solution 6 heavier.

Second move: if 125 heavier
Third move: 5 x 12
5 heavier
equal 3 lighter

Second move: if 125 equal
Third move:7 * 8

If 7 heavier : 7
if 8 heavier : 8
if equal: 4 lighter

First move if 1234 heavier, same sequence but heavy light reversed.

First move if 1234 equal;
second move 9-10 x 11 - 1

9-10 heavier (lighter)

9 x 10

9 heavier (lighter)
10 heavier (lighter)
equal: 11 lighter (heavier)

Second move 9-10 equal

Third move 12 x 1
12 lighter of heavier
 

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