Question

Write a program “hw4.c” that reads integer (less than or equal 100) from the keyboard and, on the output, writes the sum of the divisors of n (other than itself). For integers less than or equal to 1 it should print 0. For example, the
input
-3 0 1
4 5
6 12
should generate the output
0 0 0 3 1 6 16
Explanation of output:
The input -3 is less than 1, output is 0.
The input 0 is less than 1, output is 0.
The input 1 is equal to 1, output is 0.
The input 4: 4x1=4 and 2x2=4 the divisors other than itself are 2 and 1 so 2+1=3. Output is 3.
The input 5: 5x1=5 the only divisors of a prime number are itself and 1. Output is a 1.
The input 6: 6x1=6 and 3x2=6 the divisors other than itself are 1, 2 and 3. Output is 6=1+2+3.
The input 12: 12x1=12 and 6x2=12 and 3x4=12. Output=16=1+6+2+3+4.
Comment:
Primes should generate a 1. and perfect numbers should echo themselves.
(In number theory, a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. Wikipedia)
Objectives:
1. Create this routine as a function, sum_divisors() and call it from main.
2. By successively applying sum_divisors() you can compute the trajectory of the function starting with a particular value. For example, if we compute the trajectory of function on the value 1, we get 1 0 0 . . .since n is 1, sum_divisors(n) is 0, sum_divisor(sum_divisors(n)) is 0, etc. Clearly, once we see any value repeated in the sequence, we know the entire pattern. For 1, this happens after two calls to the function.
3. Prime numbers, which immediately reduce to 1 always, have a short trajectory. For example, from the value 43 we have: 43 1 0 0...which loops at 3 applications of sum_divisors().
4. Perfect numbers repeat immediately 6..6…6…6 ...so only one application of sum_divisor() detects a cycle.
5. Enhance your program so that for each integer read, it prints the number of times sum_divisor() is called in order to detect a cycle.
6. The final program( the program that should be submitted) should allow the user to enter up to 10 integers. If the user enters:
23 -4 6 45
the output should be the following:
23, 1, 0, 0 3 cycles
-4 , 0, 0 2 cycles
6, 6, 1 cycle
45, 33, 15, 9, 4, 3, 1, 0, 0 8 cycles
If the user enters an invalid input (number greater than 100 or a character) the program should end, exit(1) command ends the program.

Answer 1

#include<stdio.h>
#include<conio.h>

int sum_divisors(int n)
{
   int sum=0, i,c;
   if(n<=1)
   {
       return 0;
   }
   for(i=2;i<n;i++)
   {
       c=n%i;
       if(c==0)
       {
           sum=sum+i;
       }
   }
   return sum;
}

void main()
{
   int n[10],i,c;

   for(i=0;i<10;i++)
   {
       printf("Enter number");
       scanf("%d",&n[i]);
   }

   for(i=0;i<10;i++)
   {
       c=sum_divisors(n[i]);
       printf("\nThe input %d has sum of

divisors %d",n[i],c);

   }
   getch();

}