Problem (Combinatorics):
For any two different real numbers $x$ and $y$, we define $D(x, y)$ to be the unique integer $d$ satisfying $2^{d} \leqslant|x-y|<2^{d+1}$. Given a set of reals $\mathcal{F}$, and an element $x \in \mathcal{F}$, we say that the scales of $x$ in $\mathcal{F}$ are the values of $D(x, y)$ for $y \in \mathcal{F}$ with $x \neq y$.

Let $k$ be a given positive integer. Suppose that each member $x$ of $\mathcal{F}$ has at most $k$ different scales in $\mathcal{F}$ (note that these scales may depend on $x$ ). What is the maximum possible size of $\mathcal{F}$ ?

Solution:
We first construct a set $\mathcal{F}$ with $2^{k}$ members, each member having at most $k$ different scales in $\mathcal{F}$. Take $\mathcal{F}=\left\{0,1,2, \ldots, 2^{k}-1\right\}$. The scale between any two members of $\mathcal{F}$ is in the set $\{0,1, \ldots, k-1\}$.

We now show that $2^{k}$ is an upper bound on the size of $\mathcal{F}$. For every finite set $\mathcal{S}$ of real numbers, and every real $x$, let $r_{\mathcal{S}}(x)$ denote the number of different scales of $x$ in $\mathcal{S}$. That is, $r_{\mathcal{S}}(x)=|\{D(x, y): x \neq y \in \mathcal{S}\}|$. Thus, for every element $x$ of the set $\mathcal{F}$ in the problem statement, we have $r_{\mathcal{F}}(x) \leqslant k$. The condition $|\mathcal{F}| \leqslant 2^{k}$ is an immediate consequence of the following lemma.

Lemma. Let $\mathcal{S}$ be a finite set of real numbers, and define

$$
w(\mathcal{S})=\sum_{x \in \mathcal{S}} 2^{-r_{\mathcal{S}}(x)}
$$

Then $w(\mathcal{S}) \leqslant 1$.

Proof. Induction on $n=|\mathcal{S}|$. If $\mathcal{S}=\{x\}$, then $r_{\mathcal{S}}(x)=0$, so $w(\mathcal{S})=1$.

Assume now $n \geqslant 2$, and let $x_{1}<\cdots<x_{n}$ list the members of $\mathcal{S}$. Let $d$ be the minimal scale between two distinct elements of $\mathcal{S}$; then there exist neighbours $x_{t}$ and $x_{t+1}$ with $D\left(x_{t}, x_{t+1}\right)=d$. Notice that for any two indices $i$ and $j$ with $j-i>1$ we have $D\left(x_{i}, x_{j}\right)>d$, since

$$
\left|x_{i}-x_{j}\right|=\left|x_{i+1}-x_{i}\right|+\left|x_{j}-x_{i+1}\right| \geqslant 2^{d}+2^{d}=2^{d+1}
$$

Now choose the minimal $i \leqslant t$ and the maximal $j \geqslant t+1$ such that $D\left(x_{i}, x_{i+1}\right)=$ $D\left(x_{i+1}, x_{i+2}\right)=\cdots=D\left(x_{j-1}, x_{j}\right)=d$.

Let $E$ be the set of all the $x_{s}$ with even indices $i \leqslant s \leqslant j, O$ be the set of those with odd indices $i \leqslant s \leqslant j$, and $R$ be the rest of the elements (so that $\mathcal{S}$ is the disjoint union of $E, O$ and $R$ ). Set $\mathcal{S}_{O}=R \cup O$ and $\mathcal{S}_{E}=R \cup E$; we have $\left|\mathcal{S}_{O}\right|<|\mathcal{S}|$ and $\left|\mathcal{S}_{E}\right|<|\mathcal{S}|$, so $w\left(\mathcal{S}_{O}\right), w\left(\mathcal{S}_{E}\right) \leqslant 1$ by the inductive hypothesis.

Clearly, $r_{\mathcal{S}_{O}}(x) \leqslant r_{\mathcal{S}}(x)$ and $r_{\mathcal{S}_{E}}(x) \leqslant r_{\mathcal{S}}(x)$ for any $x \in R$, and thus

$$
\begin{aligned}
\sum_{x \in R} 2^{-r_{\mathcal{S}}(x)} & =\frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}}(x)}+2^{-r_{\mathcal{S}}(x)}\right) \\
& \leqslant \frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}_{O}}(x)}+2^{-r_{\mathcal{S}_{E}}(x)}\right)
\end{aligned}
$$

On the other hand, for every $x \in O$, there is no $y \in \mathcal{S}_{O}$ such that $D_{\mathcal{S}_{O}}(x, y)=d$ (as all candidates from $\mathcal{S}$ were in $E$ ). Hence, we have $r_{\mathcal{S}_{O}}(x) \leqslant r_{\mathcal{S}}(x)-1$, and thus

$$
\sum_{x \in O} 2^{-r_{\mathcal{S}}(x)} \leqslant \frac{1}{2} \sum_{x \in O} 2^{-r_{\mathcal{S}_{O}}(x)}
$$



Similarly, for every $x \in E$, we have

$$
\sum_{x \in E} 2^{-r_{\mathcal{S}}(x)} \leqslant \frac{1}{2} \sum_{x \in E} 2^{-r_{\mathcal{S}_{E}}(x)}
$$

We can then combine these to give

$$
\begin{aligned}
& w(S)=\sum_{x \in R} 2^{-r_{\mathcal{S}}(x)}+\sum_{x \in O} 2^{-r_{\mathcal{S}}(x)}+\sum_{x \in E} 2^{-r_{\mathcal{S}}(x)} \\
& \leqslant \frac{1}{2} \sum_{x \in R}\left(2^{-r_{\mathcal{S}_{O}}(x)}+2^{-r_{\mathcal{S}_{E}}(x)}\right)+\frac{1}{2} \sum_{x \in O} 2^{-r_{\mathcal{S}_{O}}(x)}+\frac{1}{2} \sum_{x \in E} 2^{-r_{\mathcal{S}_{E}}(x)} \\
& =\frac{1}{2}\left(\sum_{x \in \mathcal{S}_{O}} 2^{-{ }^{-} \mathcal{S}_{O}}(x)+\sum_{x \in \mathcal{S}_{E}} 2^{-r_{\mathcal{S}_{E}}(x)}\right) \quad\left(\text { since } \mathcal{S}_{O}=O \cup R \text { and } \mathcal{S}_{E}=E \cup R\right) \\
& \left.=\frac{1}{2}\left(w\left(\mathcal{S}_{O}\right)+w\left(\mathcal{S}_{E}\right)\right)\right) \quad(\text { by definition of } w(\cdot)) \\
& \leqslant 1 \quad \text { (by the inductive hypothesis) }
\end{aligned}
$$

which completes the induction.

Final answer: $2^{k}$