I am finding numbers that the sum of its divisors is either less than (deficient) equal to (perfect) or greater than (abundent) to the number. The numbers have a min and max range. I have discovered that when I use min = 1 the program works without a hitch.. (to my knowledege) But, when I use a min greater than 1 the program looses it's 1 as a divisor. Does someone have an idea how to make 1 always a factor no matter what the minimum value is? Here is my code:

Code:

#include <iostream>
#include <iomanip>
using namespace std;
int main ()
{
cout<< "What is the Minimum value? \n";
int MIN;
cin >> MIN;
cout<< "What is the Maximum value? \n";
int MAX;
cin >> MAX;
int n, div, sum;
n = MIN;
cout << "Number Factors sum classification" << endl;//Column Header
for(n; n < MAX; ++n) //Outer loop - steps through integers
{
sum = 0; //Initialize sum
cout << setw(4) << n << " "; //Row header
if (n == 1) //Special case for n == 1
{cout << "1";
sum = 1;
}
else //Otherwise
{for (div = MIN; div < MAX; ++div) //Inner loop - steps through divisors
{if (div == 1) //If the divisor equals 1 - special case
{cout << "1"; //Outputs the first factor
sum = 1; //Adds the factor to the sum
}
else if (n % div == 0) //Otherwise, checks if integer n is divisible evenly
{ //If true
if (n != div) //Checks to exclude the integer itself
{cout << ", " << div; //Outputs the next factor
sum += div; //Adds the factor to the sum
}
}
} //End inner loop
}
cout <<sum;
if(sum == n) //If the sum of the divisors equals the number
{cout << " Perfect" << endl; //Output "Perfect"
}
else if (sum < n)
{cout << " Deficient" << endl; //Output "Abundant"
}
else
{cout << " Abundant" << endl;
}
}
return 0;
}