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The 12-item balance scale puzzle is very familiar. The object is to find the lone non-standard item (if one exists) out of a group of 12 seemingly identical items, using a balance scale and a maximum of three weighings. Did you know that it is possible to accomplish the same outcome given a set of 13 seemingly identical items? I've scoured the web for a discussion of this problem/solution and have not found one. Am I the only one that has solved this problem?

BTW: If you allow four (4) weighings on a balance scale, how many seemingly identical items can you analyze and be assured of finding the lone non-standard item within the group?

If there is sufficient interest, I will publish the answers in a future post.

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    Sounds awesome, I've never heard of this kind of problem before.2012-11-27
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    I just finisher (re) working the 4 "weigh" problem. It was trickier than I remembered. Back to the 3 "weigh" problem...2012-11-28
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    I just finished (re) working the 4 "weigh" problem. It was trickier than I remembered. Back to the 3 "weigh" problem...keep in mind that a balance scale weighing yields three (3) unique outcomes. Consequently, three properly planned weighings will yield 3**3 (or 27) unique outcomes. With 13 items, any one could be heavy, any one could be light, or they could all be the same (27 outcomes). Still, there is a logistical trick needed to make it all work, and I'm not ready to reveal it just yet.2012-11-28
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    You are not the only one who has solved the 13 problem. In the 12 problem, you need not only to find the odd one out, but determine whether it is light or heavy. Unlike your question, you are given that there is exactly one non-standard item. These details are important. [This link] has a solution if you are given whether the odd one is heavy or light. One extra bit of information gives one extra bit on the outcome information. They are both good puzzles.2013-06-18

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