## Problem 3.

Find how many natural numbers $k \in\{1,2,3, \ldots, 2022\}$ have the property that, if 2022 real numbers are written on a circle such that the sum of any $k$ consecutive numbers is equal to 2022, then all 2022 numbers are equal.
Solution. Let $k \in \{1,2,3, \ldots, 2022\}$ be a number with the property that, given the real numbers $x_{1}, x_{2}, \ldots, x_{2022}$ written in a circle, the sum of any $k$ consecutive numbers is equal to 2022. For any $n \in \mathbb{N}$, we define $x_{n}=x_{r}$, where $r \in \{1,2,3, \ldots, 2022\}$, such that $n \equiv r \pmod{2022}$.

(A) For any $i \in \mathbb{N}$, we have

$$
x_{i} + x_{i+1} + \ldots + x_{i+k-1} = x_{i+1} + x_{i+2} + \ldots + x_{i+k} = 2022
$$

from which it follows that $x_{i} = x_{i+k}$

2 points

(B) If $(2022, k) = d \geq 2$, noting $x = \frac{2022 \cdot d}{k}$ and writing on the circle, in order, the numbers

$$
x, \underbrace{0,0, \ldots, 0}_{d \text{ times } - 1}, x, \underbrace{0,0, \ldots, 0, \ldots x}_{\text{d times } - 1}, \underbrace{0,0, \ldots, 0}_{\text{d times } - 1}
$$

we observe that the sum of any $k$ consecutive numbers is equal to 2022, without the numbers being equal.

Therefore, the numbers $k$ for which $(2022, k) \neq 1$ are not solutions.

(C) We show that all numbers $k \in \{1,2,3, \ldots, 2022\}$ for which $(2022, k) = 1$ are solutions. Since $(k, 2022) = 1$, there exist $u, v \in \mathbb{N}$ such that $k \cdot u = 2022 v + 1$. Then, for any $m \in \mathbb{N}$, we have:

$$
x_{m} = x_{m + k \cdot u} = x_{m + 2022 v + 1} = x_{m+1}
$$

thus all 2022 numbers are equal. 2 points

(D) In total, there are $\varphi(2022) = 672$ solutions. 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_cc89f2f54d84ee4a810cg-5.jpg?height=271&width=268&top_left_y=130&top_left_x=858)

## First selection barrier for OBMJ Iaşi, April 19, 2022  Solutions and grading criteria