You have a container containing 1 liter of pure blue liquid dye. The container is cunningly designed in such a way that up to one extra liter of liquid may be added, but only the excess over 1 liter may ever be poured out. (If the container were filled to 1.5 liters only .5 liter could be poured out.)

If you have only one liter of pure water, what is the minimum concentration you could dilute the dye in the container to?

Assume the dye is completely soluble and mixes instantaneously.

If you add 1/n of a liter to the container, the dye content changes from D to D.n/(n+1). Since you could do this operation n times, the final concentration would be (n/(n+1))^n times the initial concentration. As n goes to infinity, this goes to 1/e.

Posted by e.g. on 2006-06-08 11:03:04