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.