Sum of the digits of a number N written in all bases from 2 to N/2 in C++

In this problem, we are given a number N. Our task is to create a program to find the sum of the digits of the number N in bases from 2 to N/2.

So, we have to convert the base of the number to all bases from 2 to N/2 i.e. for n = 9, bases will be 2, 3, 4. And the find the sum of all digits in these bases.

Let’s take an example to understand the problem,


N = 5




base from 2 to N/2 is 2.
52 = 101, sum of digits is 2.

To, solve this problem, we take every number from 2 to N/2 as a base. And then to calculate the sum of digits, we will repeatedly divide the N by base i.e. N = N/base, and add the remainder value to the sum. And then add the sum values found for each base to get the result.


Program to illustrate the working of our solution 

 Live Demo

#include <iostream>
using namespace std;
int findBaseSum(int n, int base, int &sum){
   while (n > 0) {
      sum += n % base;
      n /= base;
return sum;
void CalcSumOfBaseDigits(int n, int &sum){
   for (int base = 2; base <= n / 2; base++)
      findBaseSum(n, base, sum);
int main(){
   int N = 11;
   int sum = 0;
   CalcSumOfBaseDigits(N, sum);
   cout<<"The sum of digits of "<<N<<" written in all bases from 2 to "<<(N/2)<<" is "<<sum;
   return 0;


The sum of digits of 11 written in all bases from 2 to 5 is 14