# Averaging numbers on the blackboard

Aatif sees the numbers $1 , 2 , 3 , .... , 2016$ written on the blackboard. In a move Aatif can 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 $2$?

• Incidentally, the maximum final number for values 1 .. n is $n-1 + 1/(2^n)$ – Kyle Hale Apr 12 '16 at 4:40
• @KyleHale: and the minimum final number, if I'm not mistaken, is $2 - (n+2)/2^n$. – Michael Seifert Apr 12 '16 at 13:25
• @MichaelSeifert I came up with $2 - 1/2^(n-1)$. For example, if n = 3, the minimum you can do is $7/4$. – Kyle Hale Apr 12 '16 at 17:35
• @KyleHale: Hmm, you're right — I forgot a term when I was summing things up. – Michael Seifert Apr 12 '16 at 19:40

Yes

First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$.

Now choose $2015$ and $2013$. Average = $2014$.

Choose $2014$ and $2012$. Average = $2013$.

Note that we can keep on continuing this approach and end up with a situation like $1$, $2$ and $4$ in the end.

From here, choose $2$ and $4$. Average = $3$. Average of $1$ and $3$ is $2$.

• Manal Mohania's solution works great. This is my attempt in JavaScript: (Plnkr) – hmoritz Apr 11 '16 at 17:02

Yes

Reasoning:

We can get 2 with 1 and 3. Very well then let us save 1 for last. We can get 3 with 2 and 4. Let us save 2 for that. We can get... and so on all the way up until we have saved all the numbers from 1 to 2014.

After this we...

...cannot go any higher because we cannot average 2015 and 2017; there is no 2017. However, averaging first 2014 and 2016 gives us 2015. 2015 and 2015 gives us 2015. So now we can start moving back again: 2013 and 2015 gives us 2014. 2012 and 2014 gives us 2013.

...going all the way back to...

1 and 3, which gives us 2.