# Addition of two numbers without carry

• Difficulty Level : Medium
• Last Updated : 28 May, 2022

You are given two positive numbers n and m. You have to find a simple addition of both numbers but with a given condition that there is not any carry system in this addition. That is no carry is added at higher MSBs.
Examples :

```Input : m = 456, n = 854
Output : 200

Input : m = 456, n = 4
Output : 450```

Algorithm :

```Input n, m while(n||m)
{
bit_sum = (n%10) + (m%10);

// Neglect carry
bit_sum %= 10;

// Update result
// multiplier to maintain place value
res = (bit_sum * multiplier) + res;
n /= 10;
m /= 10;

// Update multiplier
multiplier *=10;
} print res```

Approach :
To solve this problem we will need the bit by bit addition of numbers where we start adding two numbers from rightmost bit (LSB) and add integers from both numbers with the same position. Also, we will neglect carry at each position so that  carry will not affect further higher bit position.
Start adding both numbers bit by bit and for each bit take the sum of integers then neglect their carry by taking the modulo of bit_sum by 10 further add bit_sum to res by multiplying bit_sum with a multiplier specifying place value. (Multiplier got incremented 10 times on each iteration.)
Below is the implementation of the above approach :

## C++

 `// CPP program for special``// addition of two number``#include ``using` `namespace` `std;` `int` `xSum(``int` `n, ``int` `m)``{``    ``// variable to store result  ``    ``int` `res = 0;` `    ``// variable to maintain``    ``// place value``    ``int` `multiplier = 1;` `    ``// variable to maintain``    ``// each digit sum``    ``int` `bit_sum;` `    ``// Add numbers till each``    ``// number become zero``    ``while` `(n || m) {` `        ``// Add each bits``        ``bit_sum = (n % 10) + (m % 10);``        ` `        ``// Neglect carry``        ``bit_sum %= 10;``        ` `        ``// Update result``        ``res = (bit_sum * multiplier) + res;``        ``n /= 10;``        ``m /= 10;``        ` `        ``// Update multiplier``        ``multiplier *= 10;``    ``}``    ``return` `res;``}` `// Driver program``int` `main()``{``    ``int` `n = 8458;``    ``int` `m = 8732;``    ``cout << xSum(n, m);``    ``return` `0;``}`

## Java

 `// Java program for special``// addition of two number``import` `java.util.*;``import` `java.lang.*;` `public` `class` `GfG {` `    ``public` `static` `int` `xSum(``int` `n, ``int` `m)``    ``{``        ``int` `res = ``0``;``        ``int` `multiplier = ``1``;``        ``int` `bit_sum;` `        ``// Add numbers till each``        ``// number become zero``        ``while` `(``true``) {``                ` `            ``if``(n==``0` `&& m==``0``)``            ``break``;` `            ``// Add each bits``            ``bit_sum = (n % ``10``) + (m % ``10``);` `            ``// Neglect carry``            ``bit_sum %= ``10``;` `            ``// Update result``            ``res = (bit_sum * multiplier) + res;``            ``n /= ``10``;``            ``m /= ``10``;` `            ``// Update multiplier``            ``multiplier *= ``10``;``          ` `        ``}``        ``return` `res;``    ``}` `    ``// Driver function``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``8458``;``        ``int` `m = ``8732``;``        ``System.out.println(xSum(n, m));``    ``}``}``/* This code is contributed by Sagar Shukla */`

## Python3

 `# Python3 program for special``# addition of two number``import` `math` `def` `xSum(n, m) :` `    ``# variable to``    ``# store result``    ``res ``=` `0` `    ``# variable to maintain``    ``# place value``    ``multiplier ``=` `1` `    ``# variable to maintain``    ``# each digit sum``    ``bit_sum ``=` `0` `    ``# Add numbers till each``    ``# number become zero``    ``while` `(n ``or` `m) :` `        ``# Add each bits``        ``bit_sum ``=` `((n ``%` `10``) ``+``                   ``(m ``%` `10``))``        ` `        ``# Neglect carry``        ``bit_sum ``=` `bit_sum ``%` `10``        ` `        ``# Update result``        ``res ``=` `(bit_sum ``*``               ``multiplier) ``+` `res``        ``n ``=` `math.floor(n ``/` `10``)``        ``m ``=` `math.floor(m ``/` `10``)``        ` `        ``# Update multiplier``        ``multiplier ``=` `multiplier ``*` `10``    ` `    ``return` `res` `# Driver code``n ``=` `8458``m ``=` `8732``print` `(xSum(n, m))` `# This code is contributed by``# Manish Shaw(manishshaw1)`

## C#

 `// C# program for special``// addition of two number``using` `System;` `public` `class` `GfG {` `    ``public` `static` `int` `xSum(``int` `n, ``int` `m)``    ``{``        ``int` `res = 0;``        ``int` `multiplier = 1;``        ``int` `bit_sum;` `        ``// Add numbers till each``        ``// number become zero``        ``while` `(``true``) {` `            ``// Add each bits``            ``bit_sum = (n % 10) + (m % 10);` `            ``// Neglect carry``            ``bit_sum %= 10;` `            ``// Update result``            ``res = (bit_sum * multiplier) + res;``            ``n /= 10;``            ``m /= 10;` `            ``// Update multiplier``            ``multiplier *= 10;``            ``if` `(n == 0)``                ``break``;``            ``if` `(m == 0)``                ``break``;``        ``}``        ``return` `res;``    ``}` `    ``// Driver function``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 8458;``        ``int` `m = 8732;``        ``Console.WriteLine(xSum(n, m));``    ``}``}` `/* This code is contributed by Vt_m */`

## PHP

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## Javascript

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Output :

`6180`

Time Complexity: O(MAX(len(n),len(m))), where len(X) is the number of digits in a number X.

Space Complexity: O(1)

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