Daily Archives: 12 August, 2000

I was looking at the coins problem. I can see a solution
if you know the bad coin is heavier (or lighter — just
s/heavier/lighter). Here is a solution for that:

1. Split the coins into 3 groups of 4 coins.
2. Put two of the groups on the balance.
3. If they weigh the same, the third group contains the bad
   coin. Otherwise, the heavier group contains the bad coin.

4. Add two coins from one of the good groups to the bad
   group, and split those coins into 3 groups of 2.

5. perform the same weighing operation to find the group
   with the bad coin. This leaves 2 coins.

6. Weigh the last two coins. The heavier of the 2 is the
   bad coin.

This doesn’t answer the original problem, but may give
some idea of what it would look like. If you know the bad
coin is heavier and can do 3 weighs, you should be able to
pick the bad coin out of a group of 27 coins.