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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.
1
2014-06-13T11:26:46-0400
#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;

}

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