An integer number is said to be a perfect number if its fact

An integer number is said to be a perfect number if its factors, including 1 (but not the number itself), sum to the number. For example, 6 is a perfect number because 6 = 1 + 2 + 3. Write a function perfect that determines whether parameter number is a perfect number and another function list_perfect that prints its factors. Use this function in a program (main) that determines and prints all the perfect numbers between 1 and 10000. Starter code is provided along with this homework. Add perfect.c to your project and modify to get the following output.

/*Your header here */
#include <stdio.h>

int perfect(int);

void list_perfect(int);
int main()

{    

int i;    

for (i=2; i<10000; i++) {        

if (perfect(i)) {            

list_perfect(i);        }    

}    

return 0;

}

/*Returns whether a given number is perfect*/int perfect(int number){
}
/*Lists the perfect number\'s factorsExample: 6 = 1 + 2 + 3*/void list_perfect(int number){
}

Solution

#include <stdio.h>

int perfectOrNot(int number);

int main(void) {

int i,k=0;

int perfectNumberCount=0;

int perfectArray[10];

for(i=1;i<=10000;i++){

if(perfectOrNot(i))

perfectArray[k++]=i;     

perfectNumberCount+=perfectOrNot(i);            

}

printf(\"PERFECTCOUNT IS: %d\ \",perfectNumberCount);

for(k=0;k<4;k++)

printf(\"%d is a perfect number.\ \",perfectArray[k]);

return 0;

}

int perfectOrNot(int number){

int i,sum=0,perfectCount=0;                      

for(i=1;i<=(number/2);i++){            //No need to divide to numbers further than half of it                              

if(!(number%i)){                             

printf(\"%d can be divisible by %d.\ \",number,i);

sum=sum+i;

}

}

if (number==sum){

printf (\"%d is %d. PERFECT number\ \ \",number,perfectCount);

return 1;

}

else {

printf (\"%d is not a perfect number.\ \ \ \",number);

return 0;

}