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# Sum of digits of a given number to a given power

• Difficulty Level : Basic
• Last Updated : 20 Apr, 2021

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``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

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## Javascript

 ``

Output:

`13`

This article is contributed by Chinmoy Lenka. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.