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

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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

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.

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.

Then, we have to add all those mod values as the mod values obtained will represent that number in that base.

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 <bits/stdc++.h>
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.
import java.io.*;
public 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#

// C# 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

<?php
// PHP program to find sum of digits of
// n in different bases from 2 to n-1.
 
// function to calculate sum of 
// digit for a given base
function 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 = $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

<script>
 
// JavaScript program to find sum of digits of
// n in different bases from 2 to n-1.
 
    // function to calculate sum of
    // digit for a given base
    function solve(n , base) {
        // Sum of digits
        var 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) {
            var remainder = n % base;
            result = result + remainder;
            n = parseInt(n / base);
        }
 
        // returning the result
        return result;
    }
 
    function printSumsOfDigits(n) {
        // function calling for multiple bases
        for (base = 2; base < n; ++base)
            document.write(solve(n, base) + " ");
    }
 
    // Driver Code
     
        var n = 8;
        printSumsOfDigits(n);
 
// This code contributed by Rajput-Ji
 
</script>

Output

1 4 2 4 3 2

Time Complexity: O(nlogn)

Auxiliary Space: O(1)

As constant extra space is used.

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