Question

Positive integers from 1 to 45, inclusive are placed in 5 groups of 9 each. What is the highest possible average of the medians of these 5 groups?

Answer 1

31

Explanation:

First a couple of definitions:

Median is the middle value of a group of numbers.
Average is the sum of a group of numbers divided by the count of numbers.

In working this through, it becomes clear that the goal in this exercise is to increase the various medians. So how do we do that? The goal is to arrange the sets of numbers so that we have the middle values of each set be as high as possible.

For example, the highest possible median is 41 with the numbers 42, 43, 44, and 45 being higher than it and some group of four numbers being less than it. Our first set, then, consists of (with those numbers above the median in green, the median itself in blue, and those below in red):

`color(green)(45, 44, 43, 42), color(blue)(41), color(red)(x_1, x_2, x_3, x_4)`

What then is the next highest median? There needs to be five numbers between the highest median and the next one possible (four for the numbers above the median and then the median itself), so that puts us at `41-5=36`

`color(green)(40, 39, 38, 37), color(blue)(36), color(red)(x_5, x_6, x_7, x_8)`

We can do this again:

`color(green)(35, 34, 33, 32), color(blue)(31), color(red)(x_9, x_10, x_11, x_12)`

And again:

`color(green)(30, 29, 28, 27), color(blue)(26), color(red)(x_13, x_14, x_15, x_16)`

And one last time:

`color(green)(25, 24, 23, 22), color(blue)(21), color(red)(x_17, x_18, x_19, x_20)`

And it turns out that the subscripts on the `x` values can be the actual `x` values, but they need not be. They are, at this point, interchangeable.

The average of these medians is:

`(41+36+31+26+21)/5=31`