$\underline{\text { Zhukov G. }}$.

2015 natural numbers are written in a circle such that any two adjacent numbers differ by their greatest common divisor.

Find the largest natural number $N$ that is guaranteed to divide the product of these 2015 numbers.
Evaluation. Two odd numbers cannot stand next to each other, as they do not divide their even difference. Therefore, there are no fewer than half of the numbers that are even, that is, at least 1008. Since there are more than half of the numbers that are even, some two even numbers must stand next to each other. In this pair of even numbers, at least one is divisible by 4, otherwise their difference would be divisible by 4, but neither of them would be.

Suppose we have no numbers divisible by 3. Then, due to the odd number of numbers, some two adjacent numbers give the same remainder when divided by 3. These numbers divide their difference, which is divisible by 3. This is a contradiction.

Therefore, $N \geq 3 \cdot 2^{1009}$.

Example. The numbers 4, 3, 2, 1, 2, 1, .., 2, 1, 2 satisfy the condition. Their product is $3 \cdot 2^{1009}$.

## Answer

$N=3 \cdot 2^{1009}$