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 = 1

Input : 5674
Output : 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++

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// 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;
}

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Java

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// 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.

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Python

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# 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.

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C#

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// 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

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PHP

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<?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
?>

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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 :

CPP

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#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;
}

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JAVA

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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.

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Python3

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

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C#

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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

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PHP

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<?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.
?>

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Output:

9

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

This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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