# Question :

There are 10 identical vessels, one of them with a pints of water and the others empty. You are allowed to perform the following operation: take two of the vessels and split the total amount of water in them equally between them. The object is to achieve a minimum amount of water in the vessel containing all the water in the initial set up by a sequence of such operations. What is the best way to do this?

# Solution :

Averaging the amount of water in the nonempty vessel successively with each of the nine empty vessels leaves a/29 pints in it. This is the minimum amount of water achievable for that vessel. Indeed, consider m, the minimum positive amount of water among all the vessels in their current state. (Initially, m = a and our goal is to minimize it.) Since the average of two numbers is always greater than or equal to the smaller of the two, the value of m can be decreased by the averaging operation only if this operation involves a vessel containing m pints and an empty vessel. After repeating the averaging operation with each empty vessel, no empty vessel will remain, making an increase in m impossible. Hence, the ultimate minimal value of m we can get here is equal to a/29 = a/512 pints.

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