MXMDIV -EDITORIAL

editorial
,
1

Author: @somashekhar001
Tester: @somashekhar001
Editorialist: @somashekhar001

DIFFICULTY:

EASY

PREREQUISITES:

MATHS

PROBLEM:

N and M are given
a1+a2+a3+…+aN=M.
maximum possible value of the greatest common divisor of a1,a2,a3,…aN.

SOLUTIONS:

#include<bits/stdc++.h>

using namespace std;

int main(){

   int m,n;cin>>n>>m;
 
  for(int i=n;i<=m;i++){
 
    if(m%i==0){
       cout<<m/i<<endl;
      return 0;
    }
 
  }
  return 0;
}