Originally Posted by pdurrant
Four coins in just two weighings.
number the coins 1..11.
Weigh coin 3 against coin 9.
If they are the same weight, we know that they are both good coins (because they are more than three coins apart, they can't both be bad coins). So coins 1,2,10,11 must also be good coins (because the three bad coins are together). This leaves us with the five coins 4 ..8 to examine.
If they are not the same weight, we know which one is the bad coin, because it is heavier. So we now know that the bad coins are in the five coins 1..5 or 7..11.
So after one weighing of two coins we've identified five coins that must contain the three bad coins.
Weigh the two coins at the ends of the five coins (i.e. 1 and 5 or 4 and 8 or 7 and 11).
If the coins are the same weight, the bad coins are in the middle of the five. If one of them is bad, the bad coins are at that end of the row of five.
How did I do?