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Ninety nine coins plus one catch (Posted on 2014-06-01) Difficulty: 3 of 5
You are given a task of locating one faulty coin out of 99 identical in appearance coins i.e. 98 of equal weight and one of a lesser weight; and to achieve it within seven weighings, using a balance scale.

Sounds familiar and easy (37>99)?

Familiar? Yes.

Easy? Not quite!

How can you accomplish it if no coin may be weighed more than twice ?

Bonus task: After devising the requested procedure please generalize:
How many coins (n-1 normal and one lighter) can be resolved within k weighings?

No Solution Yet Submitted by Ady TZIDON    
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re(2): Any thoughts on this one? | Comment 4 of 7 |
(In reply to re: Any thoughts on this one? by Ady TZIDON)

I guess I'd prefer just a small hint.  I'd like to be able to solve it myself but I can't seem to get past the condition that no coin may be weighed more than twice. 

As you mention, this is a familiar problem and the traditional solution might require weighing a coin more than twice.  I don't see how adding that additional restriction produces a different solution.  This puzzle is not in "Tricks" so I assume there's a legitimate solution but I'm having a hard time seeing what it could be. 

  Posted by tomarken on 2014-07-02 09:46:41
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