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# 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? $\textbf{(A) }2.5\qquad\textbf{(B) }3.0\qquad\textbf{(C) }3.5\qquad\textbf{(D) }4.0\qquad \textbf{(E) }4.5$

#### Number Theory in Median

Easy
##### Suggested Book
Mathematical Circles

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  .

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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