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.
we can know whether the coin is lighter or heavier with 12 coins.
Divide into 3 groups of 4 coins, A,B and C.
Weigh A against B. If same, the fake coin is in C. Then weigh 3 coins from C against 3 of A. IF same, then the fake coin is the remaining one in C and can be determined whether lighter or heavier in one more weighing with any other coin.
If different, then we will know whether or not it is lighter or heavier as this pile contains the fake coin. then weigh one coin against another from these three. If same, then the third coin is fake and is lighter or heavier depending on what the result in weighing two showed. If different, then the coin that is lighter/heavier as per weighing two will be the fake one.
Lets say that, however, A-B were different.
Without loss of generality, we can assume that A is lighter. Label them A1,A2,A3,A4 and B1,B2,B3,B4. Now, either one of A1,A2,A3,A4 is lighter OR B1,B2,B3,B4 is heavier.
Weigh A1,A2,B1,B2 against A3,B3,C1,C2
If same, then either A4 is lighter OR B4 is heavier. This can be determined in the third weighing of A4 against a coin from C.
If different, i.e. if A1,A2,B1,B2 is heavier, then either one of B1,B2 is heavier or A3 is lighter. Then simply weigh B1 against B2 to get the result.
IF A1,A2,B1,B2 is lighter, then either one of A1,A2 is lighter or B3 is heavier. Then again weigh A1 against A2 to get the result.
This way, one can conclusively decide which coin is fake and whether it is lighter or heavier than the others.
12 coins conclusive answer
