Question: If $m=\frac{45}{2n} −1$ and both $m$ and $n$ are positive whole numbers, what is the average value of all possible values of $n$? The answer is 7?

I am not quite sure where to start. $m$ and $n$ can be any whole numbers?

Answer: 
If you are given
$$m=\frac{45}{2n-1}$$
Then you can start by multiplying both sides by $2n-1$ to get
$$(2n-1)m=45$$
Now it would be useful to list out all of the factors of $45$. We have
$$1\times45=45$$
$$3\times15=45$$
$$5\times9=45$$
$$9\times5=45$$
$$15\times3=45$$
$$45\times1=45$$
Thus the possible values of $2n-1$ are $1,3,5,9,15,$and $45$, and so the possible values of $n$ are $1,2,3,5,8,$ and $23$, whose average is $7$.
By using the [divisor function](https://en.wikipedia.org/wiki/Divisor_function) $\sigma\_x$, this can answered with a formula. The function $\sigma\_0(n)$ gives the number of divisors of $x$, and $\sigma\_1(x)$ gives the sum of the divisors.

So $\frac{\sigma\_1(n)}{\sigma\_0(n)}$ is the average of the divisors of $n$. And $\left(\frac{\sigma\_1(45)}{\sigma\_0(45)}+1\right)/2=(\frac{78}{6}+1)/2$ is the answer to the problem.