A positive integer has exactly 50 divisors. Is it possible that no difference of two different divisors is divisible by 100? Proposed by A. Chironov
Problem
Source: All-Russian MO 2024 9.2
Tags: number theory, number theory proposed
23.04.2024 06:35
04.05.2024 18:44
The answer is \boxed{\text{no}}. Assume on the contrary that no two divisors of n have a difference that is divisible by 100. Case 1: v_2(n)=0 Notice that the residue set of the divisors must be \{01,03,\dots,99\}. Since 25 is in this set 5^2|n. But 5^3 can not divide n as 5^3-5^2=100. But since n has exactly 50 divisors we have a contradiction. Case 2: v_2(n)=1 Consider the 25 divisors divisible by 2. Their residues must be \{02,06,\dots,98\}. Since 50 is in this set 5^2|n. But 5^3 can not divide n as 5^3-5^2=100. But since n has exactly 50 divisors we have a contradiction. Case 3: 4\leq v_2(n) There must be at least \frac{3}{5}\cdot 50=30 divisors of n divisible by 4. However there are only 25 possible residues, \{00,04,\dots,96\}, a contradiction.