Question

Write a program that asks the user for a lower limit and an upper limit. The program finds all pairs of amicable numbers where the first number of the pair is between those limits. The second number of the pair may or may not be between the limits. To do this sensibly, write a function int sumDivisors( int num ); that returns the sum of all the proper divisors of num.

The main program looks at each integer N between the limits. For each N, compute the sum S of its proper divisors. Then check if the sum of the divisors of S is the original number N. If N and S are the same number, then N is perfect.

The program writes out amicable pairs and perfect numbers one per line. Write out just one number if it is perfect. Write out each amicable pair just once. If both numbers of a pair are within the limits, then the pair will be found twice. So when a pair is found check that S is greater than N. If so, write out the pair. If not, then the lower number of the pair S has already been found as part of a pair.

Write sumDivisors()as a loop that checks trial divisors starting at 2 and going up until trial*trial > N. If N%trial == 0, then add trial and N/trial to the sum. Initialize the sum to 1 (since 1 divides N).

Answer 1

Editable Code:

#include <stdio.h>
int sumDivisors(int n)
{
   int trial;
   int Sum = 1;
   for (trial = 2; trial*trial < n; trial++)
   {
       if (n % trial == 0)
           Sum = Sum + trial + n / trial;
   }
   if (trial*trial == n)
       Sum = Sum + trial;
   return Sum;
}
int main()
{
   int size=0, i, j, lower_Limit, upper_Limit;
   printf("Enter the lower limit: ");
   scanf("%d", &lower_Limit);
   printf("ENter the upper limit: ");
   scanf("%d", &upper_Limit);
   size = upper_Limit - lower_Limit;
   int arr[size];
   for (i = 0; i < size; ++i)
       arr[i] = lower_Limit + i;
   for (i = 0; i < size; ++i)
   {
       for (j = 0; j < size; ++j)
       {
           if ((arr[i] == sumDivisors(arr[j])) && (arr[j] == sumDivisors(arr[i])))
               if (arr[i] == arr[j])
                   printf("The Perfect number is: %d\n", arr[i]);
               else
                   printf("The Amicable pair is: (%d %d)\n", arr[i], arr[j]);
       }
   }
   system("pause");
   return 0;
}