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 Pearls (Posted on 2002-05-23)
You have nine pearls, one of which is (as is usually the case in these problems) fake. You know that the fake pearl weighs less than the others, but it is (of course) impossible to distinguish from the others in any other way.

What is the minimum number of weighings that must be performed to find the fake pearl? How would you go about it?

 See The Solution Submitted by levik Rating: 2.7143 (7 votes)

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 SOLUTION FOR P=15 | Comment 8 of 9 |

If we had fifteen pearls at the outset and in terms of the provisions of the problem, were required to isolate the counterfeit pearl, then the required minimum number of weighings would have been three.
We take fourteen pearls out of the available fifteen pearls leaving behind a solitary pearl and place seven each of the aforementioned pearls in each pan. If the pans balance, then the solitary pearl must be fake . If they do not balance, then we take 6 out of the defective set of 7 pearls and place 3 each on the two pans. If the pans balance, then the remaining pearl must be the counterfeit one. If the pans do not balance, then isolating one pearl, we place one pearl each on the two pans. If the pans balance, the isolated pearl is the counterfeit. If the pans do not balance, then the lighter pearl on the pan is the counterfeit pearl.
THUS, THE REQUIRED MINIMUM NUMBER OF WEIGHING IN CASE OF 15 PEARLS IS INDEED 3.

 Posted by K Sengupta on 2005-11-19 00:33:47

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