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: 9876543244Input: 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; } |
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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 |
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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 |
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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 |
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Output:
9876543244
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