Question

how to write a c++ program which given an integer n, classifies it as "perfect", "abundant" or "deficient. a number is perfect if the sum of its divisor (excluding the number) is equal to the number eg. 6. a number is abundant is if the sum of its divisor (excluding the number is greater than the number eg 12?

Answer 1

Here's a function for perfect numbers:
int perfect(int number)
{
int factor=1, sum=0;
while(number>1){
if(number%factor==0){
sum+=factor;
}
factor++;
if(factor>=number)break;
}
if(sum==number)return true;
else return false;
}