I have a question, but more of a wonder, about 2014 amc8 question #24, here is 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?

(A) 2.5 (B) 3.0 (C) 3.5 (D) 4.0 (E) 4.5

So I already know the answer and HERE is what I don't know about - the solution provided on AoPS said in order to maximize the median, we need to make the first half of the numbers as small as possible, and they choose the first 49! My question is - why not choose the 50???(I have already try using 50 to solved the q's and it turns out if I use 50, I will not get the correct answer )

]]>thanks for the recommendation! I will try to remember that in the future! Also, I am kind of curious, how do you learn about math(for competition)? Do you use AoPS or any other online courses? ]]>

(before we start: remember to always put the lists in order from smallest to largest when doing medians!)

If I have a list of 5 things, say \(1,2,3,4,5\), the median is simple-- just the middle number. In this list the median is \(3\).

Actually, for all lists with an odd number of things, it is just that simple! Just take the middle number.

However, even is a little more complicated. Let's say I have 6 things: \(2,3,4,5,6,7\). The median is the middle number, but wait-- there's two middle numbers! So, the median is going to be the average of \(4\) and \(5\), which is \(4.5\).

That's how it works for all lists with an even number of things. We take the average of the middle two numbers. So, when we look at a list of 100 things, it will look something like this:

(the first \(49\) numbers) (\(50\)th number) (\(51\)st number) (last \(49\) numbers)

The middle two numbers are going to be the \(50\)th and the \(51\)st, right? So we want to make those two as big as possible so we can get the greatest median.

So to do this, let's make the first \(49\) all equal to \(1\). That means there are \(252-49=203\) cans left. We want to spread this among the remaining \(51\) as evenly as possible so the \(50\)th and the \(51\)st numbers are as big as possible. Notice how the best way to do this is making the \(50\)th equal to \(3\) and the rest equal to \(4\).

So, the \(50\)th number is \(3\) and the \(51\)st is \(4\). The median is \(3.5\).

Here's the problem with making the first \(50\) small. Remember how I split the numbers into

(the first small \(49\) numbers) (\(50\)th number) (\(51\)st number) (last \(49\) big numbers)

Well, if instead you do

(the first \(50\) small numbers) (last \(50\) big numbers)

Then the middle two numbers will still be the \(50\)th and the \(51\)st. But, the \(50\)th is small and the \(51\)st is big. This is not as optimal as making both as big as possible. Hopefully this makes sense, feel free to ask about anything!

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