EGMO 2026 Problem 2

Given a positive integer $n$, Marie plays a game where she starts with the number 1 on a blackboard. As many times as she wants, she can choose an integer $j$ such that $1 \leq j \leq n$ and replace the number $V$ on the blackboard with the number $j \cdot R\left(\frac{V}{j}\right)$. Here, $R(x)$ denotes the nearest integer to $x$; if $x$ is exactly halfway between two consecutive integers, it is rounded up. For example, $R(1.3) = 1$ and $R(1.5) = R(1.8) = 2$.

a) Prove that for each given $n$, there is a positive integer $B$ such that Marie can never end up with a number larger than $B$ on the blackboard.

b) For any given $n$, let $f(n)$ be the maximum number obtainable on the blackboard after finitely many replacements. Show that there exists a positive integer $N$ such that for all $n \geq N$, we have that 2026 divides $f(n)$.