# C Programming – HackerRank Solution | Divisor summation 1 |

### (HackerRank) Write a Modular C Programming code to solve Divisor summation 1, Given a natural number n output the summation of all its proper divisors.

Given a natural number n output the summation of all its proper divisors.

Definition: A proper divisor of a natural number is the divisor that is strictly less than the number. e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Input Format

Natural number n

Constraints

n should be positive

Output Format

Divisor summation of n

Sample Input 0

```20
```

Sample Output 0

```Divisor summation is 22
```

Sample Input 1

```-5
```

Sample Output 1

`Invalid natural number`

CODE:

```#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

void divisor(int n)
{
int sum=0,i;
for(i=1;i<n;i++)
{
if(n%i==0)
{
sum=sum+i;
}
}
printf("Divisor summation is %d\n",sum);
}

int main() {

int n;
scanf("%d",&n);
if(n<1)
{
printf("Invalid natural number\n");
exit(0);
}
else
{
divisor(n);
}
return 0;
}
```

OUTPUT

```Congratulations, you passed the sample test case.

Click the Submit Code button to run your code against all the test cases.

Input (stdin)

20

Divisor summation is 22
Expected Output

Divisor summation is 22```

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