1. The picture shows several circles connected by segments. Sasha chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:

if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with \( n \), while if they are connected, then the numbers \( a + b \) and \( n \) must have a common natural divisor greater than 1. For what smallest \( n \) does such an arrangement exist?

![](https://cdn.mathpix.com/cropped/2024_05_06_0b0e793e58d3be992fedg-01.jpg?height=274&width=277&top_left_y=800&top_left_x=907)
Answer: $n=35$.

Solution. We will make two observations.

1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then among four consecutively connected numbers, there exists a pair of adjacent numbers whose sum is not divisible by $q$. Take a chain $(a, b, c, d)$ of consecutively connected numbers. By the condition,

$$
a+d=(a+b)-(b+c)+(c+d) \vdots p
$$

Then the numbers $a$ and $d$ are also connected, which means there is a cycle of length 4 in the picture, which does not exist.

From 1) and 3), it follows that the number $n$ has at least two distinct odd prime divisors. Let there be exactly two of them (say, $p$ and $q$). We will show that they are different from 3. Suppose, for example, that $p=3$. No more than two numbers are divisible by 3 (if there are three, they form a cycle). The remaining numbers can be divided into two groups, giving remainders 1 and 2 when divided by 3. One of these groups is empty, otherwise any number from the smaller group will be connected to at least three numbers from the other group, which is impossible. The sum of the numbers in one group is not divisible by 3. Therefore, there exists a three-link chain in which the sum of any pair of connected numbers is not divisible by 3 and, hence, is divisible by $q$. But this contradicts 2).

Thus, if $n$ has exactly two distinct odd prime divisors, then $n \geqslant 5 \cdot 7=35$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>35$. The arrangement for $n=35$ is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_0b0e793e58d3be992fedg-01.jpg?height=328&width=343&top_left_y=2006&top_left_x=868)