Sum of digits of a given number to a given power

Last Updated : 17 Feb, 2023

Given a number, we need to find the sum of all the digits of a number which we get after raising the number to a specified power.
Examples:

Input: number = 5, power = 4 
Output: 13
Explanation:
Raising 5 to the power 4 we get 625.
Now adding all the digits = 6 + 2 + 5


Input: number = 9, power = 5
Output: 27
Explanation:
Raising 9 to the power 5 we get 59049.
Now adding all the digits = 5 + 9 + 0 + 4 + 9

The approach for Python is explained. we have used pow() function to calculate the base to the power value. Then we have extracted every digit as string using str() method. Since we can’t calculate the sum of strings, we converted every string digit back to integer using int() method. Finally, we used sum() function to get the sum of all the digits. This solution will look very simple in Python but it won’t be so short in other languages. After running both the codes, one can compare the time elapsed and the memory used in both the given language i.e., Python and Java.
Below is the implementation of above idea :

C++

// CPP program to illustrate the given problem 
#include<bits/stdc++.h> 
using namespace std; 
  
int calculate(int n, int power) 
{ 
    int sum = 0; 
    int bp = (int)pow(n, power); 
    while (bp != 0) { 
        int d = bp % 10; 
        sum += d; 
        bp /= 10; 
    } 
    return sum; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 5; 
    int power = 4; 
    cout << calculate(n, power); 
} 
  
// This code is contributed by Nikita Tiwari 

Java

// Java program to illustrate the given problem 
public class base_power { 
    static int calculate(int n, int power) 
    { 
        int sum = 0; 
        int bp = (int)Math.pow(n, power); 
        while (bp != 0) { 
            int d = bp % 10; 
            sum += d; 
            bp /= 10; 
        } 
        return sum; 
    } 
  
    // Driver Code 
    public static void main(String[] args) 
    { 
        int n = 5; 
        int power = 4; 
        System.out.println(calculate(n, power)); 
    } 
} 

Python3

# Python program to illustrate the given problem 
def calculate(n, power): 
    return sum([int(i) for i in str(pow(n, power))]) 
      
# Driver Code 
n = 5
power = 4
print (calculate(n, power)) 

C#

// C# program to find sum of digits of 
// a given number to a given power 
using System; 
  
public class base_power  
{ 
      
    // Function to calculate sum 
    static int calculate(int n, int power) 
    { 
        int sum = 0; 
        int bp = (int)Math.Pow(n, power); 
        while (bp != 0)  
        { 
            int d = bp % 10; 
            sum += d; 
            bp /= 10; 
        } 
        return sum; 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int n = 5; 
        int power = 4; 
        Console.WriteLine(calculate(n, power)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to illustrate 
// the given problem 
  
function calculate($n, $power) 
{ 
      
    $sum = 0; 
    $bp = (int)pow($n, $power); 
    while ($bp != 0) 
    { 
        $d = $bp % 10; 
        $sum += $d; 
        $bp /= 10; 
    } 
    return $sum; 
} 
  
// Driver Code 
$n = 5; 
$power = 4; 
echo(calculate($n, $power)); 
  
// This code is contributed by Ajit. 
?> 

Javascript

<script> 
  
// Program to illustrate the given problem 
   
   function calculate( n,  power) 
{ 
     sum = 0; 
     bp = Math.pow(n, power); 
    while (bp != 0) { 
         d = bp % 10; 
        sum =sum+ d; 
        bp = Math.floor(bp/ 10); 
    } 
    return sum; 
} 
    
// Driver Code 
  
     n = 5; 
     power = 4; 
    document.write(calculate(n, power)); 
      
// This code is contributed by simranarora5sos 
    
</script>