Caro and the three kids arrived yesterday; this morning was a little nuts. I was trying to work on my class project due next week. Between kids running around and asking questions, one broken glass, one missing cat and making breakfast, I didn't get much done. I'm having fun.
The puzzle and question (again):
You have twelve coins, one of which is a fake. It is a fake because it weighs more or less than the other eleven coins. (You do not know whether it weighs more, or less.) You have a very old balance that indicates, when coins are placed on each of the two platters, which set of coins is heavier. Can you find the fake coin in three weighings?
Answer:
Divide the 12 coins into 4 sets of 3 coins. We'll call these sets A, B, C and D. Weigh A and B. Without loss of generality, let us suppose that A and B differ in weight. Remove a coin from A, and from B; put these two coins aside. Take a coin from A and swap it with a coin from B. Reweigh the two (smaller) sets. If their weights are equal then the fake coin is one of the two you removed. If nothing changes then the fake coin is one of the coins you didn't touch. If the weight reading is reversed (lighter becomes heavier and heavier becomes lighter) then the fake coin is one of the swapped coins. For the pair of coins identified as containing a fake, weigh one of its coins against any know valid coin. If these two are of equal weight then the fake is other of the idenified pair, otherwise the fake is the coin (of the identified pair) that was just weighed.
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