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# Answer to Question #43047 in C++ for nona

Question #43047
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.
Expert's answer
``#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 != '3') && (cardNumber != '4') && (cardNumber != '5') && (cardNumber != '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; }   ``

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