We claim that only $k=1$ and $k=2$ satisfy the required property. First, if $k \geqslant 3$, we colour each integer with its residue class modulo 3 , so that, whenever two integers have the same colour, their difference is divisible by 3 , so is not a power of 2 . This shows that no $k \geqslant 3$ has the required property.

In the case $k=1$, the sequence defined by $a_{n}=2 n$ for $n=1,2, \ldots, 2023$ clearly has the required property. In the case $k=2$, we call the colours "red" and "blue", and construct, for each $n \geqslant 1$ and by induction, integers $a_{1}1$, let $a_{1}<a_{2}<\cdots<a_{n}$ be red integers (without loss of generality) having the desired property. Consider the $n+1$ integers $b_{i}=a_{n}+2^{i}$, for $i=1,2, \ldots, n+1$. If one of these, say $b_{j}$, is red, then, as $b_{j}-a_{n}=2^{j}$, the $n+1$ red integers $a_{1}<a_{2}<\cdots<a_{n}<b_{j}$ have the desired property. Otherwise, $b_{1}, b_{2}, \ldots, b_{n+1}$ are all blue, and $b_{i+1}-b_{i}=\left(a_{n}+2^{i+1}\right)-\left(a_{n}+2^{i}\right)=2^{i}$ for $i=1,2, \ldots, n$, so the $n+1$ blue integers $b_{1}<b_{2}<\cdots<b_{n+1}$ have the desired property. This completes the inductive step and hence the proof.

## BxMO 2023: Problems and Solutions

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