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.

first, divide the 10 coins into 2 piles with 5 coins each. weigh these piles against each other. the pile with the heavier coin will naturally weigh more. take this pile and divide it into 2 more piles of coins, each containing 2 coins. you will have 1 extra coin; put this aside. weigh the 2 new piles of coins against each other. if either one is heavier, weigh both coins against each other to find out the heavier coin. if both piles weigh the same, that means the 1 coin you have put aside is the heavier coin. as for the open problem, you can have more coins, as long as they are an even number.

Posted by me on 2003-06-05 22:59:02