# Sum of digits written in different bases from 2 to n-1

• Difficulty Level : Easy
• Last Updated : 08 Apr, 2021

Given a number n, find the sum of digits of n when represented in different bases from 2 to n-1.
Examples:

```Input : 5
Output : 2 3 2
Representation of 5 is 101, 12, 11 in bases 2 , 3 , 4 .

Input : 7
Output : 3 3 4 3 2```

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1. As the given question wants the sum of digits in different bases, first we have to calculate the given number of different bases and add each digit to the number of different bases.
2. So, to calculate each number’s representation we will take the mod of given number by the base in which we want to represent that number.
3. Then, we have to add all those mod values as the mod values obtained will represent that number in that base.
4. Finally, the sum of those mod values gives the sum of digits of that number.

Below are implementations of this approach

## C++

 `// CPP program to find sum of digits of``// n in different bases from 2 to n-1.``#include ``using` `namespace` `std;` `// function to calculate sum of``// digit for a given base``int` `solve(``int` `n, ``int` `base)``{``    ``// Sum of digits``    ``int` `result = 0 ;``    ` `    ``// Calculating the number (n) by``    ``// taking mod with the base and adding``    ``// remainder to the result and``    ``// parallelly reducing the num value .``    ``while` `(n > 0)``    ``{``        ``int` `remainder = n % base ;``        ``result = result + remainder ;``        ``n = n / base;``    ``}``    ` `    ``// returning the result``    ``return` `result ;``}` `void` `printSumsOfDigits(``int` `n)``{``    ``// function calling for multiple bases``    ``for` `(``int` `base = 2 ; base < n ; ++base)   ``        ``cout << solve(n, base) <<``" "``;``}` `// Driver program``int` `main()``{``    ``int` `n = 8;``    ``printSumsOfDigits(n);``    ``return` `0;``}`

## Java

 `// Java program to find sum of digits of``// n in different bases from 2 to n-1.``class` `GFG``{``// function to calculate sum of``// digit for a given base``static` `int` `solve(``int` `n, ``int` `base)``{``    ``// Sum of digits``    ``int` `result = ``0` `;``    ` `    ``// Calculating the number (n) by``    ``// taking mod with the base and adding``    ``// remainder to the result and``    ``// parallelly reducing the num value .``    ``while` `(n > ``0``)``    ``{``        ``int` `remainder = n % base ;``        ``result = result + remainder ;``        ``n = n / base;``    ``}``    ` `    ``// returning the result``    ``return` `result ;``}` `static` `void` `printSumsOfDigits(``int` `n)``{``    ``// function calling for multiple bases``    ``for` `(``int` `base = ``2` `; base < n ; ++base)``        ``System.out.print(solve(n, base)+``" "``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``8``;``    ``printSumsOfDigits(n);``}``}``// This code is contributed by Smitha`

## Python3

 `# Python program to find sum of digits of``# n in different bases from 2 to n-1.`` ` `# def to calculate sum of``# digit for a given base``def` `solve(n, base) :``     ` `    ``# Sum of digits``    ``result ``=` `0``     ` `    ``# Calculating the number (n) by``    ``# taking mod with the base and adding``    ``# remainder to the result and``    ``# parallelly reducing the num value .``    ``while` `(n > ``0``) :``    ` `        ``remainder ``=` `n ``%` `base``        ``result ``=` `result ``+` `remainder ``        ``n ``=` `int``(n ``/` `base)``     ` `    ``# returning the result``    ``return` `result`` ` `def` `printSumsOfDigits(n) :``     ` `    ``# def calling for``    ``# multiple bases``    ``for` `base ``in` `range``(``2``, n) :``        ``print` `(solve(n, base), end``=``" "``)` `# Driver code``n ``=` `8``printSumsOfDigits(n)`` ` `# This code is contributed by Manish Shaw``# (manishshaw1)`

## C#

 `// Java program to find the sum of digits of``// n in different base1s from 2 to n-1.``using` `System;` `class` `GFG``{``// function to calculate sum of``// digit for a given base1``static` `int` `solve(``int` `n, ``int` `base1)``{``    ``// Sum of digits``    ``int` `result = 0 ;``    ` `    ``// Calculating the number (n) by``    ``// taking mod with the base1 and adding``    ``// remainder to the result and``    ``// parallelly reducing the num value .``    ``while` `(n > 0)``    ``{``        ``int` `remainder = n % base1 ;``        ``result = result + remainder ;``        ``n = n / base1;``    ``}``    ` `    ``// returning the result``    ``return` `result ;``}` `static` `void` `printSumsOfDigits(``int` `n)``{``    ``// function calling for multiple base1s``    ``for` `(``int` `base1 = 2 ; base1 < n ; ++base1)``        ``Console.Write(solve(n, base1)+``" "``);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 8;``    ``printSumsOfDigits(n);``}``}``// This code is contributed by Smitha`

## PHP

 ` 0)``    ``{``        ``\$remainder` `= ``\$n` `% ``\$base` `;``        ``\$result` `= ``\$result` `+ ``\$remainder` `;``        ``\$n` `= ``\$n` `/ ``\$base``;``    ``}``    ` `    ``// returning the result``    ``return` `\$result` `;``}` `function` `printSumsOfDigits(``\$n``)``{``    ` `    ``// function calling for``    ``// multiple bases``    ``for` `(``\$base` `= 2 ; ``\$base` `< ``\$n` `; ++``\$base``)``    ``{``        ``echo``(solve(``\$n``, ``\$base``));``        ``echo``(``" "``);``    ``}``}` `// Driver code``\$n` `= 8;``printSumsOfDigits(``\$n``);` `// This code is contributed by Ajit.``?>`

## Javascript

 ``

Output :

`1 4 2 4 3 2 `

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