We have a group of 20 coins, one of which is a counterfeit coin. We know that the false coin is lighter than all of the other coins. Given a simple scale, what is the fewest number of weighings needed to determine which coins is fake?
The minimum number of weighings needed to determine the fake coin is 3. Consider the tree below, in which we follow a “branch” if that side of the scale was lighter than the other. In all cases, it will take a maximum of three weighings to determine the false coin.
[diagram included in full question set]
The Disappearing Dollar:
Three men went to a motel to get a room. The clerk told them that the price of the room was $30. Each man gave the clerk $10 to pay for the room. Later the clerk realized he had overcharged the men and gave the bellhop $5 to give back to the men. The bellhop decided that $5 was too hard to divide equally among the three men so he gave each man $1 and kept the other $2. Each man originally paid $10 and got $1 back so they paid $9 each. Three times $9 is $27 and the bellhop kept $2.
Where did the other dollar go?
There really isn't a missing dollar. The riddle fools you into thinking that what the men paid plus what the bellhop received should equal $30.
The three men paid $27, the motel received $25, and the bellhop received $2. Total paid is $27 ($9+$9+$9) and the total received is $27 ($25+$2). The books balance and there is no missing dollar.
It is only a coincidence that what the men paid plus what the bellhop received comes close to the original $30 figure. Consider if the men had initially paid $60 ($20 each) for the room and the manager gave the bellhop $35 to return to the men. The bellhop could have given each man $10 back and kept $5. Each man has now paid $10 ($30 total) the bellhop has $5. $30+$5=$35 and we see that now we're so far off from the original $60 that you don't think to look for the "missing" $25.