Answer to Question #43047 in C++ for nona

Question

2. Compute the sum of single digit numbers obtained from step 1 above:
Sum of even single-digits = 4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37

3. Compute the sum of every odd digit from right to left:
Sum of odd digits = 6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38

4. Compute the sum of the results from step 2 and step 3:
Total sum = 37 + 38 = 75

5. If the total sum from step 4 is divisible by 10, the card number is valid. Otherwise
it’s not valid.

Is 75 divisible by 10 (without remainder)? No, the card number in this
example is Invalid.

Answer 1

#include<iostream>
#include<string>
 
using namespacestd;
 
intmain() {
    cout << "Enterthe card number: ";
    string cardNumber;
    cin >>cardNumber;
 
    bool valid = true;
    int sumOfOdd = 0;
    int sumOfEven = 0;
 
    // The credit card number muststart with 3, 4, 5 or 6.
    if ((cardNumber[0] != '3') && (cardNumber[0] != '4') && (cardNumber[0] != '5') && (cardNumber[0] != '6')) {
        valid = false;
    }
 
    // The total number of digits ofa credit card number must be between 13 to 16 digits.
    if ((cardNumber.length() < 13) || (cardNumber.length() > 16)) {
        valid = false;
    }
 
    // Luhn Rule
    for (int i = 0; i <cardNumber.length(); i++) {
        char c =cardNumber[cardNumber.length() - i - 1];
        intnum = c - '0';
 
        if ((num< 0) || (num> 9)) {
            valid = false;
        }
 
        if (i % 2 == 1) {
            num = 2 *num;
            if (num> 9) {
                num =num / 10 +num % 10;
            }
            sumOfEven =sumOfEven + num;
        }
        else {
            sumOfOdd =sumOfOdd + num;
        }
    }
 
    int sum =sumOfOdd + sumOfEven;
 
    if (sum% 10 != 0) {
        valid = false;
    }
 
 
    if (valid) {
        cout << "Thecard is valid" << endl;
    }
    else {
        cout << "Thecard is invalid" << endl;
    }
    return 0;
 
}