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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(3): Any thoughts on this one? small hint - open to all | Comment 5 of 7 |
(In reply to re(2): Any thoughts on this one? by tomarken)

It is solvable with 97 COINS in 7(OR LESS) trials .

Initially I was presented with smaller numbers and erred in

"extrapolation"  trying to publish a more difficult task.

OK - Start with picking randomly 2*13 marbles and compare one set of 13  with the other,  If they balance your new problem is to continue w3ith 71 un-touched marbles within  6 or less steps..

If not, the lighter side may be seen as 6 pairs and ODD MAN OUT=compare within eachb pair..6 times at most.

Now it's up to you to continue the process and describe it in such details that I will copy it and paste as an official solution.

I hope you still like the puzzle. I certainly  do.

  Posted by Ady TZIDON on 2014-07-02 12:30:59
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