Replacing Numbers By Half Of Their Average

I claim that $n= 2^k$ for some $k \in \mathbb{Z}^+$ .

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First, I shall show that the sum of the reciprocals of the numbers written by the blackboard must decrease (not necessarily strictly) in each step. To do this, it suffices to show that $\dfrac{1}{a} + \dfrac{1}{b} \geq \dfrac{4}{a+b}$ , which is equivalent to $(a+b)^2 \geq 4ab$ , which is true.

Thus, if $t$ denotes the final number remaining after $n-1$ moves, we have $\dfrac{1}{t} \leq n \implies t \geq \dfrac{1}{n}$ . We want equality to hold. Now, we have $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{4}{a+b}$ if and only if $a=b$ . Thus, at any moment, the chosen numbers must be equal.

It follows that initially, each $1$ must be paired up with another $1$ , so $n$ must be even. Then, the configuration is equivalent to $n/2$ $1/2$ 's. Each of these $1/2$ 's must be paired up with another $1/2$ , so $n/4$ is even. We continue this way till we reach $1$ , so $n$ cannot have any non trivial odd factor, so $n$ must be a power of $2.$

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Thus, our answer is $\boxed{10}$ .

Define a 'step' to be a replacement as described in the question. Also, in my solution powers of 2 are 1, 2, 4, 8, ...

It is clear that all numbers after each step are rational.

Next we prove that all of the numbers in each step, when simplified, have a power of 2 as the denominator. In the first step, we are done as we will get $\frac {1}{2}$ and the others are '1's. Suppose it was true after the kth step. We replace two of the numbers $a, b$ with $\frac {a+b}{4}$ . Note that the denominator $d$ of $a+b$ simplified must divide the product of the denominators of $a, b$ . Thus $d$ is a power of 2. Dividing $a+b$ by 4, we can see that the denominator is still a power of 2.

So, if $n$ is a natural number such that $\frac {1}{n}$ can be achieved, it must be a power of 2.

Next we prove that all powers of 2 work. We can clearly get $\frac {1}{n}=\frac {1}{2^{0}}$ . Suppose we can get $\frac {1}{2^{k}}$ with $A$ ones in the beginning. If we have $2A$ ones, we can get two of $\frac {1}{2^{k}}$ . By applying the replacement we can get $\frac {1}{2^{k+1}}$ . By induction, all powers of 2 are possible values of n.

In conclusion the answer is the number of powers of 2 from 1 to 1000 which is 10.