Consider $n$ weights, $n \ge 2$, of masses $m_1, m_2, ..., m_n$, where $m_k$ are positive integers such that $1 \le m_ k \le k$ for all $k \in \{1,2,...,n\} $: Prove that we can place the weights on the two pans of a balance such that the pans stay in equilibrium if and only if the number $m_1 + m_2 + ...+ m_n$ is even. Estonian Olympiad
Problem
Source: 2018 Romania JBMO TST 4.4
Tags: combinatorics, weights
IV4NGM
25.03.2021 20:35
Hi, is there any solution?
trigadd123
26.03.2021 13:53
It's clear that if equilibrium can be established then the sum of the weights must be even (as we must have the same weight in both pans). Assume now that the sum of our weights is even. Place the weight of mass $m_{n}$ on one of the pans. The difference of weights between the two pans is now at most $m_{n}\leq n$. Now continue by successively placing each of the weights $m_{n-1}, m_{n-2},\dots m_{1}$ on whichever pan weighs less. Notice that after having placed the weight of mass $m_{k}$, the difference between the two pans will be at most $k$. Thus, after having placed $m_{1}$, the difference between the two pans can be at most $1$. But it can't be exactly $1$- since the sum of the weights is even and we have placed them all- and thus it must be $0$, which means that equilibrium is reached. Remark. This was the 2nd official solution, translated (and slightly adapted) from here.