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.

Hi all,
for the initial problem (So Easy)
devide 11 into 3 piles, 4,4,3 (and assuming the odd one is lighter)

1. weigh the two piles of 4s, if one is lighter==> the odd one is in it, if the are equal==> the odd one is in the pile of 3.

2. assuming the odd one is in the pile of 3, we devide the pile to 3 individual coins, the we weigh 2, if one is lighter ==> it is the one, if they are equal==> the one left is the lighter one.

3. assuming the odd one is in the pile of 4, we devide them into piles of 2s, we weigh and the lighter one has the lighter coin. finally we weigh again the 2 to see which is lighter and that is the one.
(same reasoning can be used if the odd coin is heavier)


as for the open problem:
up to 27 can be solved with the same reasoning.

1. we devide 27 into piles of 9s and weigh 2, if one is lighter==> it is the one; if equal==> the remaining pile is the one.

2. we devide the pile of 9 into 3 piles of 3s. we weigh 2 piles, if one is lighter==> it is the one; if equal==> the remaining pile is the one.

3. we devide the pile of 3 into 3 individual coins. we weigh 2 coins, if one is lighter==> it is the one; if equal==> the remaining coin is the one.

Posted by Ali on 2004-01-05 01:57:49