Problem 74

Since the average of the two smallest integers in $S$ is 5, their sum is $2 \cdot 5=10$. Since the average of the two largest integers in $S$ is 22, their sum is $2 \cdot 22=44$. Suppose that the other five integers in the set $S$ are $p<q<r<t<u$. (Note that the integers in $S$ are all distinct.) The average of the nine integers in $S$ is thus equal to $\frac{10+44+p+q+r+t+u}{9}$ which equals $6+\frac{p+q+r+t+u}{9}$. We would like this average to be as large as possible. To make this average as large as possible, we want $\frac{p+q+r+t+u}{9}$ to be as large as possible, which means that we want $p+q+r+t+u$ to be as large as possible. What is the maximum possible value of $u$? Let $x$ and $y$ be the two largest integers in $S$, with $x<y$. Since $x$ and $y$ are the two largest integers, then $u<x<y$. Since $x+y=44$ and $x<y$ and $x$ and $y$ are integers, then $x \leq 21$. For $u$ to be as large as possible (which will allow us to make $p, q, r, t$ as large as possible), we set $x=21$. In this case, we can have $u=20$. To make $p, q, r, t$ as large as possible, we can take $p=16, q=17, r=18, t=19$. Here, $p+q+r+t+u=90$. If $x<21$, then $p+q+r+t+u$ will be smaller and so not give the maximum possible value. This means that the maximum possible average of the integers in $S$ is $6+\frac{90}{9}=16$.