I made up a math problem which turned out to be not that easy to solve,
Assume we have a list of numbers from $1$ to $1000$. We are allowed to arbitrary evaluate average of numbers from the list and replace the average with the chosen numbers until we get only a one number. What is maximum and minimum of the final number?
Here I found some points that might help to solve the problem,
- If we classify the numbers so that each class has the same count of numbers, finding average of each class then averaging all (here it is only two steps to get the final number) gives $500.5$ which is the same number that can be obtained by averaging all the numbers at once. to prove this, assume we have $m$ groups and each group has $n$ numbers ($m\times n = 1000)$. The average is $\dfrac1m(\dfrac{S_1}n+\dfrac{S_2}n+\ldots+\dfrac{S_m}n)= \dfrac{S}{mn}=500.5$ . Here we had two steps I think we can generalize this for more steps of calculating averages as long as it doesn't violate the rule of classification.
- To obtain higher averages one idea is starting by least numbers. For example average of all numbers at once gives $\dfrac S{1000}$ but if we first average $1,2$ then average the $999$ numbers it gives $\dfrac {S-1.5}{999}=500.9994995$ which is higher average. On the other hand if we first average $1000,999$ we get $\dfrac{S-999.5}{999}=500.0005005$ which is smaller than the original one. So I think this approach might be a good strategy to get max and min average.
In the last example I only used two steps to get the final number, but there is no restriction in number of steps as stated in the problem.
Best Answer
This is not a full answer, but I think it is a good start.
Since average of weighted averages can be always replaced with single weighted average, in the end we always end up with some weighted average of all numbers $1, 2, \ldots 1000$. The rules put constraints on possible sets of weights, but not on which numbers are they assigned to.
Averaging one pair of numbers at a time, always using one we got in previous step we arrive at weights $$ \left\{\frac 1 2, \frac 1 4, \frac 1 8, \ldots \frac 1 {2^{998}}, \frac 1 {2^{999}}, \frac 1 {2^{999}} \right\}. $$
I think we will get highest result when using this weights for numbers in descending order (numerical result approx. 999) and lowest with this weights in ascending order (numerical result approx. 2).