Aatif averages numbers on the blackboard

Question

Aatif has averaged numbers and made the final number $2$: Averaging numbers on the blackboard

Today Aatif once again sees the numbers $ 1 , 2 , 3 , .... , 2016 $ written on the blackboard. In one move Aatif may pick any two numbers on the blackboard, erase them and write instead once their average. As an example, the numbers $1$ and $8$ may be replaced by $4 \frac{1}{2}$, and the numbers $2$ and $10$ may be replaced by $6$.

After $2015$ moves the blackboard only contains a single number. Can Aatif make his moves so that the final number is $999$?

Answer 1

The answer is :

YES

Explanation :

You can use the same technique as before to reduce number bigger than $999$ to $1001$ and numbers smaller than $999$ to $997$.

Big numbers :
First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$ Now choose $2015$ and $2013$. Average = $2014$
Choose $2012$ and $2014$. Average = $2013$ ... You can go on until you average $1000$ and $1002$ to $1001$

You go the same way for small numbers :
Choose $1$ and $3$. Average : $2$. Now take the $2$s. Their average is still $2$.
Choose $4$ and $2$. Average : $3$.
Choose $5$ and $3$, average : $4$.
You can go on until you average $998$ and $996$ in $997$.

Now you have $997$, $999$ and $1001$ on the blackboard :
Average $1001$ and $997$ in $999$.
Average $999$ and $999$ in $999$.

Generalization :

You can use the exact same technique for integers between $4$ and $2013$. Extremities are a little different but you can easily adapt this to find $2$, $3$, $2014$ or $2015$.
The final number on the blackboard can be any integer between $2$ an $2015$.