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?

The previous solution assumed that you can only resolve 7 coins when you get down to two weighings.  But when you're down to two weighings you have the traditional problem and you can resolve 9 coins.  Split the 9 into three groups of 3, and use the two weighings to find the fake coin. 

So with three weighings, you can actually resolve 9 + 5 + 5 = 19 coins.  With four weighings you can resolve 19 + 7 + 7 = 33 coins, with five weighings you can resolve 33 + 9 + 9 = 51 coins, with six weighings you can resolve 51 + 11 + 11 = 73 coins, and with seven weighings you can resolve 73 + 13 + 13 = 99 coins, just as stated in the original problem!

In general, then, within k weighings, you can resolve up to 2k^2 + 1 coins. 

Posted by tomarken on 2014-07-02 14:56:37