# Finding sum of digits of a number until sum becomes single digit

Given a number **n**, we need to find the sum of its digits such that:

If n < 10 digSum(n) = n Else digSum(n) = Sum(digSum(n))

**Examples :**

Input : 1234 Output : 1 Explanation : The sum of 1+2+3+4 = 10, digSum(x) == 10 Hence ans will be 1+0 = 1Input :5674Output :4

A **brute force** approach is to sum all the digits until sum < 10. **Flowchart:**

Below is the brute force program to find the sum.

## C++

// C++ program to find sum of // digits of a number until // sum becomes single digit. #include<bits/stdc++.h> using namespace std; int digSum(int n) { int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while(n > 0 || sum > 9) { if(n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum; } // Driver program to test the above function int main() { int n = 1234; cout << digSum(n); return 0; }

## Java

// Java program to find sum of // digits of a number until // sum becomes single digit. import java.util.*; public class GfG { static int digSum(int n) { int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || sum > 9) { if (n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum; } // Driver code public static void main(String argc[]) { int n = 1234; System.out.println(digSum(n)); } } // This code is contributed by Gitanjali.

## Python

# Python program to find sum of # digits of a number until # sum becomes single digit. import math # method to find sum of digits # of a number until sum becomes # single digit def digSum( n): sum = 0 while(n > 0 or sum > 9): if(n == 0): n = sum sum = 0 sum += n % 10 n /= 10 return sum # Driver method n = 1234 print (digSum(n)) # This code is contributed by Gitanjali.

## C#

// C# program to find sum of // digits of a number until // sum becomes single digit. using System; class GFG { static int digSum(int n) { int sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || sum > 9) { if (n == 0) { n = sum; sum = 0; } sum += n % 10; n /= 10; } return sum; } // Driver code public static void Main() { int n = 1234; Console.Write(digSum(n)); } } // This code is contributed by nitin mittal

## PHP

<?php // PHP program to find sum of // digits of a number until // sum becomes single digit. function digSum( $n) { $sum = 0; // Loop to do sum while // sum is not less than // or equal to 9 while($n > 0 || $sum > 9) { if($n == 0) { $n = $sum; $sum = 0; } $sum += $n % 10; $n = (int)$n / 10; } return $sum; } // Driver Code $n = 1234; echo digSum($n); // This code is contributed // by aj_36 ?>

## Javascript

<script> // Javascript program to find sum of // digits of a number until // sum becomes single digit. let n = 1234; //Function to get sum of digits function getSum(n) { let sum = 0; while (n > 0 || sum > 9) { if(n == 0) { n = sum; sum = 0; } sum = sum + n % 10; n = Math.floor(n / 10); } return sum; } //function call document.write(getSum(n)); //This code is contributed by Surbhi Tyagi </script>

**Output :**

1

There exists a **simple and elegant O(1) solution** for this too. The ans is given by simply :-

If n == 0 return 0; If n % 9 == 0 digSum(n) = 9 Else digSum(n) = n % 9

**How does the above logic works? **

If a number n is divisible by 9, then the sum of its digit until sum becomes single digit is always 9. For example,

Let, n = 2880

Sum of digits = 2 + 8 + 8 = 18: 18 = 1 + 8 = 9

A number can be of the form 9x or 9x + k. For the first case, answer is always 9. For the second case, and is always k.

Below is the implementation of the above idea :

## C++

#include<bits/stdc++.h> using namespace std; int digSum(int n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver program to test the above function int main() { int n = 9999; cout<<digSum(n); return 0; }

## Java

import java.io.*; class GFG { static int digSum(int n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver program to test the above function public static void main (String[] args) { int n = 9999; System.out.println(digSum(n)); } } // This code is contributed by anuj_67.

## Python3

def digSum(n): if (n == 0): return 0 if (n % 9 == 0): return 9 else: return (n % 9) # Driver program to test the above function n = 9999 print(digSum(n)) # This code is contributed by # Smitha Dinesh Semwal

## C#

using System; class GFG { static int digSum(int n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver Code public static void Main () { int n = 9999; Console.Write(digSum(n)); } } // This code is contributed by aj_36

## PHP

<?php function digSum($n) { if ($n == 0) return 0; return ($n % 9 == 0) ? 9 : ($n % 9); } // Driver program to test the above function $n = 9999; echo digSum($n); //This code is contributed by anuj_67. ?>

## Javascript

<script> function digSum(n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Driver code n = 9999; document.write(digSum(n)); // This code is contributed by code_hunt </script>

**Output:**

9

**Related Post : **

https://www.geeksforgeeks.org/digital-rootrepeated-digital-sum-given-integer/

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