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.
All the previous posted solutions make a second weighing based on the results of the first. But it is possible to make two predetermined weighings.
Partition the coins into four groups A,B,C,D with A<C, A+B=C, A+C=D. Then make weighings A+B vs C and A+C vs D.
The fake coin is heavy when:
the left side is heavy both times (in group A)
left side is heavy first, then balance (in group B)
right side heavy first, left side heavy second (in group C)
or balance first then right side heavy second (in group D).
The fake coin is light for each cases' converse, which the other four cases:
the right side is heavy both times (in group A)
right side is heavy first, then balance (in group B)
left side heavy first, right side heavy second (in group C)
or balance first then left side heavy second (in group D).
Another solution
