# Clusters on the Number Line: Official Solution

*English translation by SOTA – AI Community of the Russian original. Organisers who would like this translation removed can email sota.ai.community@gmail.com.*

*Task 1 of the school stage (Moscow) of the All-Russian School Olympiad (VsOSh) 2025/26 in artificial intelligence, grades 9–11: official answer and solution (the statement is in a separate file). Original: [sol-ai-9-11-sch-msk-25-26.pdf](https://vos.olimpiada.ru/upload/files/Arhive_tasks/2025-26/school/ai/sol-ai-9-11-sch-msk-25-26.pdf).*

**Answer:** 2, 3, 7

**Scoring criterion:** exact match of the answer — 12 points

**Maximum score for the task — 12**

**Solution.**

Consider the numbers 21, 22, 28. If the whole set splits into exactly two clusters, then at least two of these three lie in the same cluster, so the difference of one of the pairs is divisible by $k$. The possible differences are $1, 6, 7$. Since $k \geq 2$, we have

$$
k \mid 6 \quad \text{or} \quad k \mid 7,
$$

that is, the remaining candidates are $k \in \{2, 3, 6, 7\}$.

Check:

- $k = 2$: all the numbers are even or odd $\Rightarrow$ two clusters.
- $k = 3$: all the numbers leave a remainder of 0 or 1 $\Rightarrow$ two clusters.
- $k = 7$: all the numbers leave a remainder of 0 or 1 $\Rightarrow$ two clusters.
- $k = 6$: already the numbers $42 \equiv 0$, $1 \equiv 1$, $21 \equiv 3$, $22 \equiv 4 \pmod 6$ give four different remainders $\Rightarrow$ not two clusters.

Thus, only $k = 2, 3, 7$ work.