Problem:
One day the Beverage Barn sold cans of soda to customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
Answer Choices:
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Solution:
Suppose the numbers of cans purchased by the customers are listed in increasing order. The median is the average of the th and st numbers in the ordered list. To maximize the median, minimize the first numbers by taking them all to be . If the th number is , then the sum of all numbers would at least , which is too large. If instead the th number is and the following numbers all equal , then the sum of the numbers is and the median is .
To maximize the median, the largest should be the same and as large as possible. The lower should be as small as possible. The median of the list will be the average of the th and st numbers. If every customer has can of soda, there are left to distribute. Giving the upper three more each gives the top four cans each total) and the lower one each total). There are cans left. Giving the th person the extra means the th has cans, and the st has cans for a median of .
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions