Code check for bad credit pset1

Question

This is my attempt for the bad credit problem in pset1 and I'm having trouble debugging my code. Any help would be appreciated.

#include <stdio.h>
#include <cs50.h>

bool checksum(long long number);

int main(void)
{
    long long number,i;
    int count, first_dig, first_two;


    printf ("Number: ");
    number = get_long_long();

    for (count = 1, i = number; i >= 100; count++)
    {
        i = i/10;
    } 

    first_two = i;
    first_dig = i/10;
    ++count;

    if (count < 13 || count > 16 || count == 14)
    {
         printf("INVALID\n");
    }

    if ((count == 13 || count == 16) && first_dig == 4 && checksum(number) == true)
        printf ("VISA\n");

    else if ( (count == 15) && (first_two == 34 || first_two == 37) && checksum(number) == true )
        printf ("AMEX\n");

    else if( (count == 16) && (first_two == 51 || first_two == 52 || first_two == 53 || first_two == 54 || first_two == 55 ) && checksum(number) == true)
         printf ("MASTERCARD\n");

    else printf ("INVALID\n");

}  

bool checksum(long long number)
{
    long long value = number;
    int i, j, digits[20], sum = 0;

    for (i = 0; value > 0; ++i)
    {
        digits[i] = value % 10;
        value /= 10;

    }


    for (j = 0; j < i; j++)
    {
        if ( (j % 2) == 0)
        {
             sum += digits[j];
        }

        else sum += (digits[j] * 2);
    }

    if (sum % 10 == 0)
    {
        return true;
    }

    else return false;
}   

Answer 1

sum += (digits[j] * 2);

is not how the algorithm works. For digits 0..4, it's that, but for 5..9, you add the two digits of the result, so it's

sum += (digits[j] * 2) - 9;

you can combine that for example to

sum += digits[j] * 2 - (digits[j] > 4) * 9;

or

sum += digits[j] * 2 - (digits[j] > 4 ? 9 : 0);

or

sum += (digits[j] % 5) * 2 + digits[j] / 5;

or some different variation