Find the last two missing digits of the given phone number

Given eight digits of a phone number as an integer N, the task is to find the missing last two digits and print the complete number when the last two digits are the sum of given eight digits.
Examples:

Input: N = 98765432 
Output: 9876543244
Input: N = 10000000 
Output: 1000000001 

Approach:

• Get the eight digits of the phone number from N one by one using the Modulo 10 operator (%10).
• Add these digits in a variable say sum to get the sum of the eight digits.
• Now, there are two cases: 
  • If sum < 10 then it is a single digit i.e. insert 0 in the beginning to make it a two digit number without affecting the value.
  • Else sum is the number represented by the last two digits.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach

#include <iostream>
using namespace std;

// Function to find the last two
// digits of the number and
// print the complete number
void findPhoneNumber(int n)
{

    int temp = n;
    int sum;

    // Sum of the first eight
    // digits of the number
    while (temp != 0) {
        sum += temp % 10;
        temp = temp / 10;
    }

    // if sum < 10, then the two digits
    // are '0' and the value of sum
    if (sum < 10)
        cout << n << "0" << sum;

    // if sum > 10, then the two digits
    // are the value of sum
    else
        cout << n << sum;
}

// Driver code
int main()
{
    long int n = 98765432;

    findPhoneNumber(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to find the last two
// digits of the number and
// print the complete number
static void findPhoneNumber(int n)
{

    int temp = n;
    int sum = 0;

    // Sum of the first eight
    // digits of the number
    while (temp != 0)
    {
        sum += temp % 10;
        temp = temp / 10;
    }

    // if sum < 10, then the two digits
    // are '0' and the value of sum
    if (sum < 10)
        System.out.print(n + "0" + sum);

    // if sum > 10, then the two digits
    // are the value of sum
    else
        System.out.print(n +""+ sum);
}

// Driver code
public static void main(String[] args)
{
    int n = 98765432;

    findPhoneNumber(n);
}
}

// This code is contributed by PrinciRaj1992

Python 3

# Python 3 implementation of the approach

# Function to find the last two
# digits of the number and
# print the complete number
def findPhoneNumber(n):
    temp = n
    sum = 0

    # Sum of the first eight
    # digits of the number
    while (temp != 0):
        sum += temp % 10
        temp = temp // 10

    # if sum < 10, then the two digits
    # are '0' and the value of sum
    if (sum < 10):
        print(n,"0",sum)

    # if sum > 10, then the two digits
    # are the value of sum
    else:
        n = str(n)
        sum = str(sum)
        n += sum
        print(n)

# Driver code
if __name__ == '__main__':
    n = 98765432

    findPhoneNumber(n)

# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to find the last two
// digits of the number and
// print the complete number
static void findPhoneNumber(int n)
{
    int temp = n;
    int sum = 0;

    // Sum of the first eight
    // digits of the number
    while (temp != 0)
    {
        sum += temp % 10;
        temp = temp / 10;
    }

    // if sum < 10, then the two digits
    // are '0' and the value of sum
    if (sum < 10)
        Console.Write(n + "0" + sum);

    // if sum > 10, then the two digits
    // are the value of sum
    else
        Console.Write(n + "" + sum);
}

// Driver code
static public void Main ()
{
    int n = 98765432;

    findPhoneNumber(n);
}
}

// This code is contributed by jit_t

Javascript

<script>

// Javascript implementation of the approach

// Function to find the last two
// digits of the number and
// print the complete number
function findPhoneNumber(n)
{

    let temp = n;
    let sum=0;

    // Sum of the first eight
    // digits of the number
    while (temp != 0) {
        sum += temp % 10;
        temp = Math.floor(temp / 10);
    }

    // if sum < 10, then the two digits
    // are '0' and the value of sum
    if (sum < 10)
        document.write(n + "0" + sum);

    // if sum > 10, then the two digits
    // are the value of sum
    else
        document.write(n + "" + sum);
}

// Driver code

    let n = 98765432;

    findPhoneNumber(n);

// This code is contributed by Mayank Tyagi

</script>

Output:

9876543244

Time Complexity: O(log₁₀n)

Auxiliary Space: O(1)