# Problem of the Week Problem C and Solution Average Out

## Problem

Four different positive integers have a mean (average) of $$100$$. If the positive difference between the smallest and largest of these integers is as large as possible, determine the average of the other two integers.

Extra Problem: Can you interpret the following picture puzzle?

## Solution

Let $$a$$,$$b$$, $$c$$, and $$d$$ represent four distinct positive integers such that $$a<b<c<d$$.

Since the average of the four positive integers is $$100$$, their sum can be determined by multiplying their average by $$4$$. Therefore, the sum of the numbers is $$4\times 100=400$$. That is, $$a+b+c+d=400$$.

For the difference between the largest integer and the smallest integer to be as large as possible, we want the smallest integer, $$a$$, to be as small as possible. The smallest positive integer is $$1$$, so $$a=1$$.

Since the sum of the four positive integers is $$400$$ and the smallest integer, $$a$$, is $$1$$, the sum of the remaining three integers is $$b+c+d=400-1=399$$.

For the difference between the largest integer and the smallest integer to be as large as possible, we also want the largest integer, $$d$$, to be as large as possible. For $$d$$ to be as large as possible, $$b$$ and $$c$$ must be as small as possible. The two positive integers, $$b$$ and $$c$$, must be different and cannot equal $$1$$, since $$a=1$$. Therefore, $$b=2$$ and $$c=3$$, the smallest two remaining distinct positive integers. It follows that $$d$$, the largest of the four positive integers, is $$399-2-3=394$$. (This was not required but has been provided for completeness.)

The average of the middle two positive integers, $$b$$ and $$c$$, is $$\frac{2+3}{2}=\frac{5}{2}=2.5$$.

Extension:

How would your answer change if it was also required that the average of $$b$$ and $$c$$ was an integer greater than or equal to $$3$$?

Extra Problem Answer: Slightly above average.