Program to check for a Valid IMEI Number

International Mobile Equipment Identity (IMEI) is a number, usually unique, to identify mobile phones, as well as some satellite phones. It is usually found printed inside the battery compartment of the phone, but can also be displayed on-screen on most phones by entering *#06# on the dialpad, or alongside other system information in the settings menu on smartphone operating systems. The IMEI number is used by a GSM network to identify valid devices and therefore can be used for stopping a stolen phone from accessing that network. 

The IMEI (15 decimal digits: 14 digits plus a check digit) includes information on the origin, model, and serial number of the device. 

The IMEI is validated in following steps:

1. Starting from the rightmost digit, double the value of every second digit (e.g., 7 becomes 14).
2. If doubling of a number results in a two digits number i.e greater than 9(e.g., 7 × 2 = 14), then add the digits of the product (e.g., 14: 1 + 4 = 5), to get a single digit number.
3. Now take the sum of all the digits.
4. Check if the sum is divisible by 10 i.e.(total modulo 10 is equal to 0) then the IMEI number is valid; else it is not valid.

Example:

Input IMEI : 490154203237518

Output : Since, 60 is divisible by 10, hence the given IMEI number is Valid.

Implementation:

C++

// C++ program to check whether the
// given EMEI number is valid or not.
#include<bits/stdc++.h>
using namespace std;

// Function for finding and returning
// sum of digits of a number
int sumDig(int n)
{
    int a = 0;
    while (n > 0)
    {
        a = a + n % 10;
        n = n / 10;
    }
    return a;
}

bool isValidIMEI(long n)
{

    // Converting the number into
    // String for finding length
    string s = to_string(n);
    int len = s.length();

    if (len != 15)
        return false;

    int sum = 0;
    for(int i = len; i >= 1; i--)
    {
       int d = (int)(n % 10);

       // Doubling every alternate digit
       if (i % 2 == 0)
           d = 2 * d;

       // Finding sum of the digits
       sum += sumDig(d);
       n = n / 10;
    }

    return (sum % 10 == 0);
}

// Driver code
int main()
{
    // 15 digits cannot be stored
    // in 'int' data type
    long n = 490154203237518L;

    if (isValidIMEI(n))
        cout << "Valid IMEI Code";
    else
        cout << "Invalid IMEI Code";

    return 0;
}

// This code is contributed by Yash_R

Java

// Java program to check whether the
// given EMEI number is valid or not.
import java.io.*;
class IMEI
{
    // Function for finding and returning
    // sum of digits of a number
    static int sumDig(int n)
    {
        int a = 0;
        while (n > 0)
        {
            a = a + n % 10;
            n = n / 10;
        }
        return a;
    }

    static boolean isValidIMEI(long n)
    {
        // Converting the number into String
        // for finding length
        String s = Long.toString(n);
        int len = s.length();

        if (len != 15)
            return false;

        int sum = 0;
        for (int i = len; i >= 1; i--)
        {
            int d = (int)(n % 10);

            // Doubling every alternate digit
            if (i % 2 == 0)
                d = 2 * d;

            // Finding sum of the digits
            sum += sumDig(d);
            n = n / 10;
        }

        return (sum % 10 == 0);
    }

    // Driver code
    public static void main(String args[]) throws IOException
    {
        // 15 digits cannot be stored in 'int' data type
        long n = 490154203237518L;

        if (isValidIMEI(n))
            System.out.println("Valid IMEI Code");
        else
            System.out.println("Invalid IMEI Code");

    }
}

Python3

# Python3 code to check whether the
# given EMEI number is valid or not

# Function for finding and returning
# sum of digits of a number
def sumDig( n ):
    a = 0
    while n > 0:
        a = a + n % 10
        n = int(n / 10)

    return a

# Returns True if n is valid EMEI
def isValidEMEI(n):

    # Converting the number into
    # String for finding length
    s = str(n)
    l = len(s)

    # If length is not 15 then IMEI is Invalid
    if l != 15:
        return False

    d = 0
    sum = 0
    for i in range(15, 0, -1):
        d = (int)(n % 10)
        if i % 2 == 0:

            # Doubling every alternate digit
            d = 2 * d

        # Finding sum of the digits
        sum = sum + sumDig(d)
        n = n / 10
    return (sum % 10 == 0)

# Driver code
n = 490154203237518
if isValidEMEI(n):
    print("Valid IMEI Code")
else:
    print("Invalid IMEI Code")

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to check whether the
// given EMEI number is valid or not.
using System;

class GFG {

    // Function for finding and
    // returning sum of digits
    // of a number
    static int sumDig(int n)
    {
        int a = 0;
        while (n > 0)
        {
            a = a + n % 10;
            n = n / 10;
        }

        return a;
    }

    static Boolean isValidIMEI(long n)
    {

        // Converting the number into
        // String for finding length
        String s = n.ToString();
        int len = s.Length;

        if (len != 15)
            return false;

        int sum = 0;
        for (int i = len; i >= 1; i--)
        {
            int d = (int)(n % 10);

            // Doubling every alternate
            // digit
            if (i % 2 == 0)
                d = 2 * d;

            // Finding sum of the digits
            sum += sumDig(d);
            n = n / 10;
        }

        return (sum % 10 == 0);
    }

    // Driver code
    public static void Main()
    {

        // 15 digits cannot be stored in
        // 'int' data type
        long n = 490154203237518L;

        if (isValidIMEI(n))
            Console.Write("Valid IMEI Code");
        else
            Console.Write("Invalid IMEI Code");

    }
}

// This code is contributed by parashar.

Javascript

<script>
    // javascript program to check whether the
    // given EMEI number is valid or not.

    // Function for finding and returning
    // sum of digits of a number
    function sumDig(n)
    {
        let a = 0;
        while (n > 0)
        {
            a = a + n % 10;
            n = parseInt(n / 10, 10);
        }
        return a;
    }

    function isValidIMEI(n)
    {

        // Converting the number into
        // String for finding length
        let s = n.toString();
        let len = s.length;

        if (len != 15)
            return false;

        let sum = 0;
        for(let i = len; i >= 1; i--)
        {
          let d = (n % 10);

          // Doubling every alternate digit
          if (i % 2 == 0)
              d = 2 * d;

          // Finding sum of the digits
          sum += sumDig(d);
          n = parseInt(n / 10, 10);
        }

        return (sum % 10 == 0);
    }


    // 15 digits cannot be stored
    // in 'int' data type
    let n = 490154203237518;

    if (isValidIMEI(n))
        document.write("Valid IMEI Code");
    else
        document.write("Invalid IMEI Code");

// This code is contributed by vaibhavrabadiya117.
</script>

Output

Valid IMEI Code

Time complexity : O(n log₁₀ n)
Auxiliary Space: O(n)