Programming C - Debugging Numbers-Strings Conversion Problems

Question

I'm working on an exercise problem for a programming class and I'm a bit out of my depth debugging it. My main problem is that I have had to convert strings into integers and vice-versa and I don't understand well how that works. I'll attach a copy of the exercise and another of my code. Any and all help is appreciated.

CODE:

#include <cs50.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

string Company_Check(long number);
bool Luhn (long number);

int main(void)
{
    //Prompt for card Number
    long card = get_long("Number:");

    //Applying Luhn's Algorithm
    bool validity = Luhn(card);
    if (validity==true) //The card passed the test
    {
        //Finding and printing company name
        printf("\nCompany: %s\n", Company_Check(card));
    } else //The card didn't pass the test
    {
        printf("INVALID (Luhn’s Algorith)\n");
    }
}

string Company_Check(long number)
{
    //Declaring variables
    char num[30];
    char char_first_two[3]; //This is where I'll save the first two digits of the number
    int int_first_two;
    char char_first_one[3]; //This is where I'll save the first one digit of the number
    int int_first_one;


    //Creating the two 'First digits' var
    int first_two = 10*(atoi(num[0]))+ atoi([num[1]);
    int fist_one = atoi(num[0]);

    //Analizing cases
    if (first_two == 34 || first_two == 37)
    {
        printf("American Express\n");
    } else if (first_two == 51 || first_two == 52 || first_two == 53 || first_two == 54 || first_two == 55)
    {
        printf("Mastercard\n");
    } else{
        //Could be (1) Visa or (2) INVALID
        if (first_one == 4)
        {
            printf("Visa\n");
        } else
        {
            printf("INVALID\n");
        }
    }
}

bool Luhn (long number) //Done
{
    int i; //Used in the for loops
    int sum1; //Used to tally the sums in Step 4
    int sum2; //Used to tally the sums in Step 4
    int total; //Used to tally the sums in Step 4
    int cod; //Counter of Digits
    int lastnumber; //Used in Steps 5 and 6
    char num[30]; //Card number & total in String format
    char fot[30]; //First Every-Other number list
    char sot[30]; //Second Every-Other number list

    //1. Converting the long NUMBER to a string NUM
    sprintf(num, "%d", number);
    //2. Calculating the number of digits of the number
    cod = 0;
    for(i=0; num[i+1]!= '\0' ; i++)
    {
        cod++;
    }

    //3. Creating two CHAR arrays of the first and second every-other number list

    if (cod%2==0)//number of digits is even
    {
        for(i=0; num[i+1]!= '\0' ; i++)
        {
            if (i%2==0) //We're dealing with the FIRST set of every other numbers
            {
                fot[i/2]=num[i];
            }
            else //We're dealing with the SECOND set of every other numbers
            {
                sot[((int)(i/2))+1]=num[i];
            }
        }
        fot[(cod/2)+1]= '\0';
        sot[(cod/2)+1]= '\0';
    } else  //number of digits is odd
    {
        for(i=0; num[i+1]!= '\0' ; i++)
        {
            if (i%2==0) //We're dealing with the SECOND set of every other numbers
            {
                sot[i/2]=num[i];
            }
            else //We're dealing with the FIRST set of every other numbers
            {
                fot[((int)(i/2))+1]=num[i];
            }
        }
        sot[(cod/2)+1]= '\0';
        fot[(cod/2)+1]= '\0';
    }

    //4. Sum Algorithm
    sum1=0;
    sum2=0;
    total=0;
    for(i=0 ; fot[i+1]!= '\0' ; i++)
    {
        sum1 =+ atoi(fot[i]);
    }
    for(i=0 ; sot[i+1]!= '\0' ; i++)
    {
        sum2 =+ atoi(sot[i]);
    }
    total = sum1 + sum2;

    //5. Isolating the last number
    //converting the total into a string (using var NUM because its no longer needed)
    sprintf(num, "%d", total);
    //Calculating the number of digits
    cod = 0;
    for(i=0; num[i+1]!= '\0' ; i++)
    {
        cod++;
    }
    //putting the last number in a var
    lastnumber = atoi(num[cod-1]);

    //6. Studying cases
    if (lastnumber==0)
    {
        return true;
    }
    else
    {
        return false;
    }
}

Problem:

Problem