Program for credit card number validation

Write a program that prompts the user to enter a credit card number as a long integer and Display whether that card is valid or invalid.

Credit card numbers follow certain patterns.
A credit card number must have between 13 and 16 digits. It must start with:

  • 4 for Visa cards
  • 5 for Master cards
  • 37 for American Express cards
  • 6 for Discover cards

The problem can be solved by using Luhn algorithm.
Luhn check or the Mod 10 check, which can be described as follows (for illustration,
consider the card number 4388576018402626):

Step 1. Double every second digit from right to left. If doubling of a digit results in a
two-digit number, add up the two digits to get a single-digit number (like for 12:1+2, 18=1+8).

Step 2. Now add all single-digit numbers from Step 1.
4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37

Step 3. Add all digits in the odd places from right to left in the card number.
6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38

Step 4. Sum the results from Step 2 and Step 3.
37 + 38 = 75

Step 5. If the result from Step 4 is divisible by 10, the card number is valid; otherwise, it is invalid.
Examples :

Input : 379354508162306
Output : 379354508162306 is Valid

Input : 4388576018402626
Output : 4388576018402626 is invalid

Java

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// Java program to check if a given credit
// card is valid or not.
import java.util.Scanner;
  
public class CreditCard {
    // Main Method
    public static void main(String[] args)
    {
        long number = 5196081888500645L;
  
        System.out.println(number + " is "
        (isValid(number) ? "valid" : "invalid"));
    }
  
    // Return true if the card number is valid
    public static boolean isValid(long number)
    {
       return (getSize(number) >= 13 && 
               getSize(number) <= 16) && 
               (prefixMatched(number, 4) || 
                prefixMatched(number, 5) || 
                prefixMatched(number, 37) || 
                prefixMatched(number, 6)) && 
              ((sumOfDoubleEvenPlace(number) + 
                sumOfOddPlace(number)) % 10 == 0);
    }
  
    // Get the result from Step 2
    public static int sumOfDoubleEvenPlace(long number)
    {
        int sum = 0;
        String num = number + "";
        for (int i = getSize(number) - 2; i >= 0; i -= 2
            sum += getDigit(Integer.parseInt(num.charAt(i) + "") * 2);
          
        return sum;
    }
  
    // Return this number if it is a single digit, otherwise,
    // return the sum of the two digits
    public static int getDigit(int number)
    {
        if (number < 9)
            return number;
        return number / 10 + number % 10;
    }
  
    // Return sum of odd-place digits in number
    public static int sumOfOddPlace(long number)
    {
        int sum = 0;
        String num = number + "";
        for (int i = getSize(number) - 1; i >= 0; i -= 2
            sum += Integer.parseInt(num.charAt(i) + "");        
        return sum;
    }
  
    // Return true if the digit d is a prefix for number
    public static boolean prefixMatched(long number, int d)
    {
        return getPrefix(number, getSize(d)) == d;
    }
  
    // Return the number of digits in d
    public static int getSize(long d)
    {
        String num = d + "";
        return num.length();
    }
  
    // Return the first k number of digits from 
    // number. If the number of digits in number
    // is less than k, return number.
    public static long getPrefix(long number, int k)
    {
        if (getSize(number) > k) {
            String num = number + "";
            return Long.parseLong(num.substring(0, k));
        }
        return number;
    }
}

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C#

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// C# program to check if a given 
// credit card is valid or not.
using System;
  
class CreditCard {
      
    // Main Method
    public static void Main()
    {
        long number = 5196081888500645L;
        Console.Write(number + " is "
                     (isValid(number) ?
                     "valid" : "invalid"));
    }
  
    // Return true if the card number is valid
    public static bool isValid(long number)
    {
    return (getSize(number) >= 13 && 
            getSize(number) <= 16) && 
            (prefixMatched(number, 4) || 
            prefixMatched(number, 5) || 
            prefixMatched(number, 37) || 
            prefixMatched(number, 6)) && 
            ((sumOfDoubleEvenPlace(number) + 
            sumOfOddPlace(number)) % 10 == 0);
    }
  
    // Get the result from Step 2
    public static int sumOfDoubleEvenPlace(long number)
    {
        int sum = 0;
        String num = number + "";
        for (int i = getSize(number) - 2; i >= 0; i -= 2) 
            sum += getDigit(int.Parse(num[i] + "") * 2);
          
        return sum;
    }
  
    // Return this number if it is a 
    // single digit, otherwise, return 
    // the sum of the two digits
    public static int getDigit(int number)
    {
        if (number < 9)
            return number;
        return number / 10 + number % 10;
    }
  
    // Return sum of odd-place digits in number
    public static int sumOfOddPlace(long number)
    {
        int sum = 0;
        String num = number + "";
        for (int i = getSize(number) - 1; i >= 0; i -= 2) 
            sum += int.Parse(num[i] + "");     
        return sum;
    }
  
    // Return true if the digit d
    // is a prefix for number
    public static bool prefixMatched(long number, int d)
    {
        return getPrefix(number, getSize(d)) == d;
    }
  
    // Return the number of digits in d
    public static int getSize(long d)
    {
        String num = d + "";
        return num.Length;
    }
  
    // Return the first k number of digits from 
    // number. If the number of digits in number
    // is less than k, return number.
    public static long getPrefix(long number, int k)
    {
        if (getSize(number) > k) 
        {
            String num = number + "";
            return long.Parse(num.Substring(0, k));
        }
        return number;
    }
}
  
// This code is contributed by nitin mittal.

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Output:

5196081888500645 is valid

This article is contributed by Vishal Kumar Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal



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