The original puzzle There are 9 coins of identical appearance and one must find the faulty one, which is heavier than the others, by using a balance
only twice.
The Solution This puzzle is simple but not that evident. It is a ternary problem and we are misled into a binary frame of mind by the sight of the balance, with its 2 platters, as is the case in Three Lights. One must remember that the balance can do 3 things, tilt either way, or remain level. The coins should be divided in 3, not in 2, the faulty coin being in those left over when the balance remains level. 1. Place 3 on each side, the faulty coin being on the side that tilts, or in those left over. 2. Place 1 on each side, the faulty coin being on the side that tilts, or the one left over.
Pedagogy You will find an application of this in the Devices Chapter.
A wicked variation There is another version which is far more difficult and whose solution will not be divulged. There are 12 coins of identical
appearance and one must find the faulty one, which may be either heavier or lighter than the others, by using a balance only
three times.
Let us know how you make out with this one !
