# 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 :**

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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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