Recursive digit sum hacker rank solution in c

Problem:

https://www.hackerrank.com/challenges/recursive-digit-sum/

Solution is very simple for this problem but since the value of n can be 10^1000000 which cannot be stored in a variable of any data type in c, we need to store each digit in n in a string.

13 points solution is as follows: Using the fact that 

 recursive sum of digits is 9 if number is multiple of 9, else n % 9

#include<stdio.h>

int main()

{

        unsigned long long int n,k;

        scanf("%llu%llu",&n,&k);

        int ans=((n%9)*(k%9))%9;  // ans=(n*k)%9

        if(ans)

            printf("%llu",ans);

        else

            printf("9");

}

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int superDigit(char* n, int k) {
       int k1,i;
    k1=k%9;
    if(k1 == 0)
        return 9;
    else
    {
        int sum=0;
        for(i=0;n[i] != '\0';i++)
            sum=(sum+n[i]-48)%9;
        if((sum*k)%9 == 0)
            return 9;
        else
            return (sum*k)%9;
     }
}

int main() {
    char* n = (char *)malloc(512000 * sizeof(char));
    int k;
    scanf("%s %i", n, &k);
    int result = superDigit(n, k);
    printf("%d\n", result);
    return 0;
}