A set of 101 coins consists of 100 genuine coins of equal weight plus one fake coin which is either
a. lighter
OR
b. heavier
than a genuine one.
Using only equal-arm-balance, please determine in two weighings whether it is a. or b.
IMHO it is a nice puzzle, but I doubt whether the result has any practical meaning.
Objections? Let my know.
Source: Leningrad Math. Competition, former USSR.
Split the coins into three groups A B and C; with the constraints A=B and A<=C<=A+B.
Compare A vs B.
Case 1: A and B balance.
Then all coins in A and B are real. Then compare C vs an equal number of coins taken from A and B. C being heavy or lighter will tell if the fake coin is real or light.
Case 2: A and B do not balance.
Then all the coins in C are real. Then compare A vs an equal number of coins taken from C. If that balances then the fake coin was in B and if it doesn't balance the fake was in A. Then the first weighing gives us the info to tell if the fake is light or heavy.
I did not specify a specific number of coins, just some constraints. So getting into solid numbers we can have (A,B,C) = {(33,33,35), (32,32,37), (31,31,39), (30,30,41), (29,29,43), (28,28,45), (27,27,47), (26,26,51)} and the methodology applied will be valid.
Solution
