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DaveC426913

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Yes. But your algorithm for finding the heavy ballis as likely to require 6561 weighings, which is also thevaishakh said:

So let's rephrase:

What method is

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DaveC426913

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Split the pile into three groups of 2187 balls each. Call these piles 1a, 1b, and 1c. Weigh 1a versus 1b. If they don't balance, the heavier set contains the heavy ball. If they do balance, set 1c contains the heavy ball. Discard the two sets that don't contain the heavy ball. Repeat 6 more times, by which time only 3 balls will be left. The final (eighth) weighing will tell which is the heavy ball.

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I got it now men, 6561= 3^8.

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Further the manager found after few days that the same fault has creeped into a number of machines. Still the weighing has to be used once to find all the faulty machines. Find the method.

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I hope that [itex]n[/itex] is a huge number and the scale is very accurate.vaishakh said:

Further the manager found after few days that the same fault has creeped into a number of machines. Still the weighing has to be used once to find all the faulty machines. Find the method.

For problem 1, take 1 ball from machine #1, 2 from #2, ..., 1000 from #1000. Those balls would mass to 500.5 kg if the machines worked right. The excess mass over 500.5 kg by 0.1 g is the index of the faulty machine.

For problem 2, take 1 ball from machine #1, 2 from #2, 4 from #3, ..., 2^999 from #1000, for a total of 2^1000 - 1 grams. Divide the excess weight by 0.1 g and express the answer in base 2. All the one digits in the answer represent faulty machines.

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