$2024$ ones and $2024$ twos are arranged in a circle in some order. Is it always possible to divide the circle into a) two (contiguous) parts with equal sums? b) three (contiguous) parts with equal sums? Proposed by Fedir Yudin
Problem
Source: Kyiv City MO 2024 Round 2, Problem 7.3
Tags: combinatorics, construction
MS_Kekas
10.02.2024 06:31
Seems like this problem destroyed the best mathematicians out here... Understandable...
Tintarn
10.02.2024 14:18
OK, here you go.
Yes, this is always possible. Note that we just want a block with sum $3036$. Indeed, start behind a $1$ and go as long as the sum is at most $3036$. After that the sum will be $3035$ or $3036$. In the second case, take this block, in the first case, add the previous $1$.
No this is not always possible. Indeed, take the scenario in which ones and twos alternate. We want to have three blocks with sum $2024$. But it's easy to see that this is only possible with a block starting and ending with a two (just check mod $3$). But of course we cannot find three such blocks.
bever209
31.03.2024 00:53
My solution for a is the same as @above so I will just post my solution for b: [Block of $1011$ $1$s][Block of $1012$ $2$s][Block of $1$ $1$][Block of $1012$ $2$s] We clearly want the lone $1$ to be in a block with some other $1$, but to do that it needs to cross $1012$ $2$s which is not possible because the sum will be more than $2024$.