CAPS Match 2024

We will show these numbers exist. For that we define sequences $a_1,a_2,\dots ,a_{2024}$ and $b_1,b_2,\dots ,b_{2024}$ and then consider numbers $c_i=2^{a_i}\cdot 5^{b_i}$ for $i=1,2,\dots ,2024$. We choose the sequences $a_i$ and $b_i$ in such a way that $a_i$ is increasing, $b_i$ is decreasing, and the differences $a_i-a_j$ and $b_j-b_i$ were all mutually distinct for all indices $i>j$. This will be enough because $$\frac{c_i}{c_j}=\frac{2^{a_i}\cdot 5^{b_i}}{2^{a_j}\cdot 5^{b_j}}=\frac{2^{a_i-a_j}}{5^{b_j-b_i}},$$ this number has a decimal expansion of a length $b_j-b_i$, whereas analogously, $\frac{c_j}{c_i}$ has a length of $a_i-a_j$.

We now construct the needed sequences, starting with $a_i$. We will do it inductively. Take $a_1=1,a_2=2$. When we have the numbers $a_1,a_2,\dots ,a_i$, then by choosing $a_{i+1}=2a_i$ we will achieve $a_{i+1}-a_i>a_i-a_1$, therefore all newly added differences will be higher than the previous ones.

We can construct $b_i$ similarly, starting at the end by taking $b_{2024}=a_{2024}$, then $b_{2023}=2b_{2024}$, and so on. Since $b_{2023}-b_{2024}=a_{2024}$, all the differences in $b_i$ will be at least $b_{2023}-b_{2024}=a_{2024}$.

Remark: In our construction, $a_i=2^{i-1}$ and $b_i=2^{4049-i}$.