Find how many positive integers $k\in\{1,2,\ldots,2022\}$ have the following property: if $2022$ real numbers are written on a circle so that the sum of any $k$ consecutive numbers is equal to $2022$ then all of the $2022$ numbers are equal.
Problem
Source: Romania JBMO TST 2022
Tags: combinatorics, Romanian TST
Iora
04.05.2022 20:43
There were no solutions in this post, so I though I could share my 'solution'
Answer: All the positive odd integers $k$
Proff:
Let these positive integers be $p_1,p_2,...,p_{2022}$ , observe that:
$$p_1+p_2+...+p_k=p_2+p_3+...+p_{k+1} \Longrightarrow p_1= p_{k+1}$$After repeating this prosses $2022$ times, the prosses will be repeated $\ mod 2022$, basically after some time, we will come back to $p_1,p_2$.
When $k$ is even, the indexes of $p_i$ will be either odd or even ( if $i$ is odd, odd, if $i$ is even, even) Hence the indexes will never be opposite reverse ( even-odd), hence the prosses will repeat and we can only conclude that same indexes equals to each other (even=even, odd=odd).
But when $k$ is odd, the indexes change their odd-even property, hence at some time, with $ \mod 2022$ all the $p_i$ will be equal to each other, so anwer is all odd $k$
Is the solution right? I would appreciate criticism ( ofc it is not)
Slave
04.05.2022 21:03
Iora wrote: There were no solutions in this post, so I though I could share my 'solution'
Answer: All the positive odd integers $k$
Proff:
Let these positive integers be $p_1,p_2,...,p_{2022}$ , observe that:
$$p_1+p_2+...+p_k=p_2+p_3+...+p_{k+1} \Longrightarrow p_1= p_{k+1}$$After repeating this prosses $2022$ times, the prosses will be repeated $\ mod 2022$, basically after some time, we will come back to $p_1,p_2$.
When $k$ is even, the indexes of $p_i$ will be either odd or even ( if $i$ is odd, odd, if $i$ is even, even) Hence the indexes will never be opposite reverse ( even-odd), hence the prosses will repeat and we can only conclude that same indexes equals to each other (even=even, odd=odd).
But when $k$ is odd, the indexes change their odd-even property, hence at some time, with $ \mod 2022$ all the $p_i$ will be equal to each other, so anwer is all odd $k$
Is the solution right? I would appreciate criticism I think you're wrong. For instance observe $k=3$. I guess that the answer is all $k$ such that $\gcd (k,2022)=1$.
Iora
12.05.2022 20:37
bumping this, can someone give a hint?
green_leaf
12.05.2022 21:02
All $k$ satisfying $\gcd(k,2022)=1$ works. Since we have $a_i+a_{i+1}+ \cdots +a_{i+k-1}=2022$ and $a_{i+1}+ a_{i+2}+ \cdots a_{i+k-1}+a_{i+k}=2022$, subtracting them gives $a_i=a_{i+k}$. If $(k, 2022)=1$, it is easy to see that all numbers will be equal. Now suppose $k$ is not coprime to $2022$. Then divide the numbers into groups such that two numbers $a_i$ and $a_j$ are in the same group only if $i \equiv j+tk\pmod{2022}$. Let each group consist of equal numbers such that the sum of the number in every group is 2022. Observe that every consecutive group of $k$ numbers contain distinct numbers. Thus the sum of every consecutive group of $k$ numbers will be $2022$ and thus this is a valid construction of unequal numbers satisfying the given property for $(k,2022)>1$ and we are done.
Pashtet671
17.08.2024 19:43
Hey, @green_leaf , maybe it's easier to prove that all numbers are equal if $(k, 2022) = 1$ defining "addition" operation for indices of a? I don't really know how to formulate it strictly, but $a_i = a_{i+1}$ is the same as $a_i = a_{i+k} = a_{i+2k} = ... = a_{i+1}$, so $x*k \equiv 1$ (mod 2022). Which is, of course, only possible if $gcd(k, 2022) = 1$