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?
Solution:
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?
Solution:
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.