Given a number N in decimal base, find the sum of digits in any base B

Given a number N in decimal base, the task is to find the sum of digits of the number in any base B.

Examples:

Input: N = 100, B = 8
Output: 9
Explnantion:
(100)8 = 144
Sum(144) = 1 + 4 + 4 = 9

Input: N = 50, B = 2
Output: 3
Explnantion:
(50)2 = 110010
Sum(110010) = 1 + 1 + 0 + 0 + 1 + 0 = 3

Approach: Find unit digit by performing modulo operation on number N by base B and updating the number again by N = N / B and update sum by adding the unit digit at each step.



Below is the implementation of above approach

C++

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// C++ Implementation to Compute Sum of
// Digits of Number N in Base B
  
#include <iostream>
using namespace std;
  
// Function to compute sum of
// Digits of Number N in base B
int sumOfDigit(int n, int b)
{
      
    // Initilize sum with 0
    int unitDigit, sum = 0;
    while (n > 0) {
          
        // Compute unit digit of the number
        unitDigit = n % b;
  
        // Add unit digit in sum
        sum += unitDigit;
  
        // Update value of Number
        n = n / b;
    }
    return sum;
}
  
// Driver function
int main()
{
    int n = 50;
    int b = 2;
    cout << sumOfDigit(n, b);
    return 0;
}

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Java














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// Java Implementation to Compute Sum of 


// Digits of Number N in Base B 


class GFG 


{ 


  


// Function to compute sum of 


// Digits of Number N in base B 


static int sumOfDigit(int n, int b) 


{ 


      


    // Initilize sum with 0 


    int unitDigit, sum = 0; 


    while (n > 0)  


    { 


          


        // Compute unit digit of the number 


        unitDigit = n % b; 


  


        // Add unit digit in sum 


        sum += unitDigit; 


  


        // Update value of Number 


        n = n / b; 


    } 


    return sum; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int n = 50; 


    int b = 2; 


    System.out.print(sumOfDigit(n, b)); 


} 


} 


  


// This code is contributed by PrinciRaj1992 



















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Python3














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# Python3 Implementation to Compute Sum of 


# Digits of Number N in Base B 


  


# Function to compute sum of 


# Digits of Number N in base B 


def sumOfDigit(n, b): 


  


    # Initilize sum with 0 


    unitDigit = 0


    sum = 0


    while (n > 0): 


          


        # Compute unit digit of the number 


        unitDigit = n % b 


  


        # Add unit digit in sum 


        sum += unitDigit 


  


        # Update value of Number 


        n = n // b 


  


    return sum


  


# Driver code 


n = 50


b = 2


print(sumOfDigit(n, b)) 


  


# This code is contributed by ApurvaRaj 



















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C#














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// C# Implementation to Compute Sum of  


// Digits of Number N in Base B  


using System; 


  


class GFG  


{  


  


// Function to compute sum of  


// Digits of Number N in base B  


static int sumOfDigit(int n, int b)  


{  


      


    // Initilize sum with 0  


    int unitDigit, sum = 0;  


    while (n > 0)  


    {  


          


        // Compute unit digit of the number  


        unitDigit = n % b;  


  


        // Add unit digit in sum  


        sum += unitDigit;  


  


        // Update value of Number  


        n = n / b;  


    }  


    return sum;  


}  


  


// Driver code  


public static void Main(String[] args)  


{  


    int n = 50;  


    int b = 2;  


    Console.Write(sumOfDigit(n, b));  


}  


} 


  


// This code is contributed by PrinciRaj1992 



















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Output:


3







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