You have 3 unmarked pitchers that hold 6 cups, 10 cups, and 15 cups. The 10 cup pitcher is filled with lemonade, but you have to dilute it so people at the party don't complain that it's too sour. Unfortunately, I have, in the name of logic, broken your only measuring cup.

How can you dilute the lemonade to 20 cups such that all of the lemonade is of equal concentration? You have a sink with unlimited water, but you can't waste any lemonade. Try using as few moves as possible.

Fill the 6 cup pitcher (the "6") from the "10"
Pour the contents of the "6" into the "15"
Again, fill the "6" from the "10" -- you now have 4 cups in the "6" an empty "10" and 6 cups in the "15"
Top off the "6" with water and pour the entire contents into the "15" -- you now have empty "6" and "10", and 10 cups lemonade and 2 cups water in the "15". Stir.

Fill the "6" from the "15" -- you now have 5 cups lemonade and 1 cup water in each of the "6" and the "15", and an empty "10"
Pour the contents of the "6" into the "10" and top off with water -- the "10" now has 5 cups lemonade and 5 cups water, which is the dilution we want.
Fill the "6" again from the "15", emptying the "15" in the process.
Pour the contents of the "10" into the "15".
Once again, Pour the contents of the "6" into the "10" and top off with wanter -- the "10" now has 5 cups of lemonade and 5 cups of water, as does the "15". Stir both, and serve to your guests.

I count twelve moves, where a move is any transfer of liquid either from pitcher to pitcher or from the sink into a pitcher, but I don't see any reason to believe this is the fewest moves possible.

Posted by Paul on 2005-07-26 02:17:51