2014 Canadian MO Problems/Problem 5
Problem
Fix positive integers and . A list of n integers is written in a row on a blackboard. You can choose a contiguous block of integers, and I will either add to all of them or subtract from all of them. You can repeat this step as often as you like, possibly adapting your selections based on what I do. Prove that after a finite number of steps, you can reach a state where at least of the numbers on the blackboard are all simultaneously divisible by .
Solution
We aim to prove that given positive integers written in a row on a blackboard, with , we can achieve a state where at least numbers are simultaneously divisible by after a finite number of steps. For any number on the blackboard, consider its value modulo , denoted . Initially, each number can take one of the values modulo . The allowed operations, which involve adding or subtracting to all numbers in a contiguous block, cyclically adjust the residues of the selected block within . This gives us complete control over the residues of the selected block modulo .
Our goal is to ensure that at least numbers are divisible by , which corresponds to ensuring that their residues modulo are . To do this, note that at every step, we can select a contiguous block of numbers and adjust their residues to reduce the number of distinct residues present among the numbers. Specifically, we aim to minimize the number of residues different from . If fewer than numbers are divisible by , then more than numbers must have non-zero residues.
By the pigeonhole principle, at least one residue must appear more than once. Using our operations, we can choose a contiguous block of numbers containing all occurrences of (or a subset) and adjust their residues modulo to reduce towards .
Repeating this process for different residues as necessary, we can iteratively increase the number of numbers divisible by until at least numbers satisfy this property. The process must terminate after a finite number of steps because each operation reduces the sum of the absolute values of the residues modulo across all numbers. Since this sum is non-negative and decreases strictly with each operation, the process cannot continue indefinitely. Therefore, after a finite number of steps, we can ensure that at least numbers are divisible by .
~sitar