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

The problem statement includes printing the sum of digits of N, which will be the user input, when written in different bases from 2 to N−1.

In this problem, we will be provided any positive integer N and we need to represent that number in a different base numeral system from 2 to N−1 and find the sum of the digit of each different base numeral system.

In the base−n numeral system, every digit of the representation of any number in that numeral system from right represents the number of times power of n from 0 to 31.

For example, when 5 is being represented in base−2 numeral system,

It is represented as 101 i.e $\mathrm{2^{2}+2^{0}=5}$

Meanwhile, when 5 is being represented in base−3 numeral system,

It is represented as 12 $\mathrm{3^{1}+2*3^{0}=3+2=5}$

Let’s understand the problem with the examples below.

Input

N=6

Output

2 2 3 2

Explanation − The input given is 6. So, here we need to find the sum of digits of N i.e. 6 when being represented in a different base numeral system from 2 to 5.

When 6 is represented in base−2 numeral system, it is written as 110 i.e.$\mathrm{2^{2}+2^{1}=6}$

The sum of digits of representation of 6 in base−2 is 2

The representation of 6 in base−3 numeral system is 20 i.e $\mathrm{2^{2}*3^{1}=6}$.The sum of digits of 6 in base−3 is 2.

The representation of 6 in base−4 numeral system is 12 i.e.$\mathrm{4^{1}+2*4^{0}=4+2=6}$ The sum of the digits of 6 in base−4 representation is 1+2=3.

The representation of 6 in base−5 numeral system is 11 i.e $\mathrm{5^{1}+5^{0}=6}$. The sum of digits of 6 in the base−5 numeral system is 2.

Hence, our required output is 2 2 3 2.

Input

N=9

Output

2  1  3  5  4  3  2

Explanation − Since the input is 9, we need to represent 9 in a different base numeral system from 2 to 8 and find the sum of digits of each representation. Therefore,

9 in base−2 : 1001 thus the sum of digits is 2.

9 in base−3 : 100, thus the sum of digits is 1.

9 in base−4 : 21, making the sum of digits 3.

9 in base−5 : 14, making the sum of digits 5.

9 in base−6 : 13, making the sum of digits 4.

9 in base−7 : 12, making the sum of digits 3.

9 in base−8 : 11, making the sum of digits 2.

Hence, our required output is 2 1 3 5 4 3 2.

Let’s understand the algorithm to print the sum of digits of a number when represented in a different base numeral system.

Algorithm

As we need to find the sum of digits of representation of a number N for each base from 2 to N−1, we will create a function to find the sum of N for every base representation.

To represent any number N in any base number, we take the mod of the number with the base number which represents each digit from right and then divide the number by the base number until N becomes 0 to get the representation of that number in the particular base numeral system.

For example we need to represent 9 in the base−5 numeral system.

Dividing 9 by 5 gives 4 as a remainder, so the rightmost digit will be 4 and then update N by dividing 9 by 5 which gives 1.

Now, dividing 1 by 5 gives 1 as a remainder, so the next digit to the left of the previous digit will be 1 and then N will be 0.

So the representation of 9 in base−5 numeral system will be 14 which represent $\mathrm{5^{1}+4*5^{0}=9}$

Since we just need the sum of digits of representation of N in different base values, we will simply keep adding remainders of N when divided by base value and update N by dividing it by base value until N is greater than 0 to get the sum of the digits of N when written in different base values.

We will use this algorithm to calculate the sum of digits of N when represented in different base values from 2 to N−1 in our approach in order to solve the problem.

Approach

The steps to follow to implement the above algorithm in our approach:

• We will simply make a function to give the sum of digits of N when represented in different base values from 2 to N−1, where the passed parameters will be N and the base number.

• We will iterate in a while loop until N is greater than 0 where we will find the remainder of N when divided by base number and add it in the variable to store the sum of digits and then update N by dividing N by base number.

• Once N becomes 0, return the sum of digits stored in the variable.

• Iterate in a for loop from i=2 to i<=N−1 to find the sum of digits of N when represented in different base numbers, where i represents a particular base number.

• We will print the sum of digits of each different bases by calling the function to calculate the sum of digits of N when written in different bases in the for loop for every value of i from 2 to N−1.

Example

// C++ program to print the Sum of digits written in different bases from 2 to N-1
#include<bits/stdc++.h>

using namespace std;

//function to give the sum of digits of N when represented in a particular base
int sumOfDigit(int N,int B)
{
    int sum = 0;  //variable to store the sum of digits
    //iterate until N is greater than zero
    while(N > 0)
    {
        
     int rem=N % B;  //finding the remainder of N when divided by base number 
     
     sum=sum + rem;  //add remainder with sum  to store the sum of digits
     
     N=N / B;  //update N by dividing N by base number
     
    }
    
    return sum;  //return the sum of digits stored in the variable
}
int main()
{
    int N=10;
    
    cout<<"Sum of digits written in different bases from 2 to N-1 : " <<endl;
    
    //using for loop from i=2 to i<=N-1 to find the sum of digits of N when represented in different base number
    for(int i =2; i<=N-1; i++){
        
        //printing the sum of digits of each different bases by calling the function to calculate the sum of digits of N 
        cout<<sumOfDigit(N,i)<<"  ";
        
    }
    return 0;
}

Output

Sum of digits written in different bases from 2 to N-1 
2  2  4  2  5  4  3  2 

Time complexity− O(N*logN) , because it took log N time to calculate the sum of digits in different bases.

Space complexity − O(1) , because no extra space is used in our approach.

Conclusion

The concept of printing the sum of digits of N when represented in different bases from 2 to N−1 was discussed in the article. We used an efficient approach to find the sum of digits of N in different base values in C++ in (log N) runtime and within constant space.

I hope you understand the concept and the solution to the problem after reading this article.