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:
- Split the coins into 3 groups of 4 coins.
- Put two of the groups on the balance.
- If they weigh the same, the third group contains the bad
coin. Otherwise, the heavier group contains the bad coin.
- Add two coins from one of the good groups to the bad
group, and split those coins into 3 groups of 2.
- perform the same weighing operation to find the group
with the bad coin. This leaves 2 coins.
- Weigh the last two coins. The heavier of the 2 is the
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.