2014 AMC 8 Problem 24 Number Theory in Median

 Understand the Problem

One day the Beverage Barn sold $252$ cans of soda to $100$ 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? $\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$
Source of the problem

2014 AMC 8 Problem 24

Topic

Number Theory in Median

Difficulty Level
Easy
Suggested Book
Mathematical Circles

Start with hints

Do you really need a hint? Try it first!

Recall the concept of Median . https://en.wikipedia.org/wiki/Median
 As here 252 cans of soda sold to  even no of customers i.e.  to 100 customers , so median will be the average of the 50th and 51st largest amount of cans per  person .  
First they each have one can atleast one can . Now think for the rest 152 cans and 50th and 51st customers . Then in order to proceed divide 100 customers in two groups as first 49 customers and second 51 customers .
To minimize the first 49, they will  be sold no more cans. Now 152 cans left to divide among next 51 customers . Taking \( \frac {152}{51} \) gives us 2 and a remainder of 50. Seeing this, the largest number of cans the 50th person could have is 3 ( including the can in 1st distribution), which leaves 4( including the can in 1st distribution)  to the rest of the people. The average of 3 and 4 is 3.5  . So the maximum possible median is 3.5  . 

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