Question

Hi, I'm trying to write C code that will print the odd abundant numbers between 1 and 5000 and I've written a program for it but it doesn't work. If you would be able to help me fix it or point out where I went wrong that would be awesome.

# include <stdio.h>

int main () {
  
int getSum(int n)
int i, n, j, sum = 0;
  
/* print initial statement to give context of program*/
  
printf(" The odd abundant numbers between 1 and 5000 are: \n");
scanf("%d", &n);
  
/* first declare a range for the integer i*/
  
for (i=1; 1<=n; i++); {

/* determine each factor for every integer i*/

for (j = 1; j <= i/2; j++) {
if (i % j == 0) {
sum = sum + j;
}
}
/* print abundant number alone */
  
if (sum > i)
  
printf("%d ", n);

sum = 0;
}
/* determine whether integer i is odd*/
  
if (i % 2 != 0) {
  
/* print all integers that satisfy these requirements*/
  
printf("%d", n); }
  
printf("\n");

return 0;
}

  
  

Answer 1

Your code was not properly formed, and there were alot of errors, So I recreated the code. Below is the working code

#include<stdio.h>
#include<math.h>
int checkIfNumberAbundant(int num)
{
int sum = 0;
int i;
for (i=1; i<=sqrt(num); i++)
{
if (num%i==0)
{
if (num/i == i)
sum = sum + i;
  
else
{
sum = sum + i;
sum = sum + (num / i);
}
}
}
sum = sum - num;
return (sum > num);
}
  
void main()
{
int count = 0;
int i = 0;
for(i = 0; i < 5000;i++) {
if(i%2 != 0) {
if(checkIfNumberAbundant(i) > 0) {
printf("%d ",i);
count++;
}
}
}
printf("Total number of abundant numbers less than 5000 is %d",count);
}