In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.

Using only a balance beam for only three times, show how you can determine the 'odd' coin.

Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.

nikki is right, g(2) would have to equal 3, g(1) would have to equal 0, g(3)=12, g(4)=39, g(5)=120
as you can tell, i have figured out the formula, but i want to see if you can figure it out now that you have more information

Posted by Jonathan on 2003-03-17 15:07:20