Numbers on a Blackboard, Extreme Edition

A common idea to prove these problems (where no. of ways is 1):

In general let $a \# b = \frac{4ab-1}{4(a+b-1)}$ (this is the definition of the operation #). Now all that is sufficient to prove uniqueness of end result is to show commutativity, $a \# b = b \# a$ (evident) and associativity, $a \# (b \# c) = (a \# b) \# c$ (some algebra). Now any order of applying operations on any pairs of the 2015 numbers will always yield the same result after 2014 turns.

Now to find that last remaining number you can simply choose any systematic way to apply the operations that makes induction convenient.

Of course, this may not be the elegant solution (like finding the invariant), especially if you need to find the remaining number). However, it is the most flexible as it applies to most of these type of questions (note that if an invariant can be found, the operation is commutative and associative).

Two numbers $a,b$ are replaced by $\frac{4ab-1}{4(a+b-1)}$ and so, if $x,y > 0$ , the numbers $\tfrac12 + x^{-1}$ and $\tfrac12 + y^{-1}$ are replaced by $\tfrac12 + (x+y)^{-1}$ . This proves the associativity of the replacement rule.

Since the $2015$ initial numbers are all equal to $\tfrac12 + 2^{-1}$ , and $2\times2015 = 4030$ , it is clear that the final number must be $\tfrac12 + 4030^{-1} = \tfrac{1008}{2015}$ , no matter the order in which the numbers are removed.

[This is a hint.]

Consider a second blackboard where every number $x$ on the first blackboard is replaced with the number $\dfrac{1}{2x-1}$ on the second blackboard.

Does the rule change into something more easy to work with?