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Recursive sum of digit in n^x, where n and x are very large

  • Difficulty Level : Easy
  • Last Updated : 08 Apr, 2021

Given very large numbers n and x, we need to find the sum of digits of n^x such that :
 

If n^x < 10    
    digSum(n^x) = n^x
Else         
    digSum(n^x) = Sum(digSum(n^x))

Examples: 
 

Input :  5  4
Output :  4
We know 54 = 625
Sum of digits in 625 = 6 + 2 + 5 = 13
Sum of digits in 13 = 1 + 3 = 4

Input :  546878 56422
Output :  7

 

Prerequisite : Recursive sum of digits
The idea is: 
Sum of digits repeats after every 6th exponents. 
Let SoD(n) = a 
Let b = x % 6 
then 
SoD(n^x) = SoD(a^b) except for b = 1 when a = 3 & 6 
SoD(n^x) = 9 forall x > 1, when a = 3 & 6 
 

C++




// CPP Code for Sum of
// digit of n^x where
// n and x are very large
#include <bits/stdc++.h>
using namespace std;
  
// function to get sum
// of digits of a number
     long digSum(long n)
    {
        if (n == 0)
            return 0;
        return (n % 9 == 0)
                ? 9 : (n % 9);
    }
  
    // function to return sum
     long PowDigSum(long n, long x)
    {
         
        // Find sum of digits in n
        long sum = digSum(n);
  
        // Find remainder of exponent
        long rem = x % 6;
  
        if ((sum == 3 || sum == 6)
                         && x > 1)
            return 9;
  
        else if (x == 1)
             return sum;
  
        else if (x == 0)
             return 1;
  
        else if (rem == 0)
             return digSum((long)pow(sum, 6));
  
        else
            return digSum((long)pow(sum, rem));
    }
  
// Driver code
int main()
{
   int n = 33333;
   int x = 332654;
   cout << PowDigSum(n, x);
    return 0;
}
 
// This code is contributed by Gitanjali.

Java




// Java Code for
// Sum of digit of
// n^x where n and
// x are very large
import java.util.*;
 
class GFG {
 
    // function to get sum
    // of digits of a number
    static long digSum(long n)
    {
        if (n == 0)
            return 0;
        return (n % 9 == 0) ? 9 : (n % 9);
    }
 
    // function to return sum
    static long PowDigSum(long n, long x)
    {
        // Find sum of digits in n
        long sum = digSum(n);
 
        // Find remainder of exponent
        long rem = x % 6;
 
        if ((sum == 3 || sum == 6) && x > 1)
            return 9;
 
        else if (x == 1)
            return sum;
 
        else if (x == 0)
            return 1;
 
        else if (rem == 0)
            return digSum((long)Math.pow(sum, 6));
 
        else
            return digSum((long)Math.pow(sum, rem));
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
 
        int n = 33333;
        int x = 332654;
 
        System.out.println(PowDigSum(n, x));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3




# Python3 Code for Sum
# of digit of n^x
import math
 
# function to get
# sum of digits of
# a number
def digSum(n):
    if (n == 0):
        return 0
    if n % 9==0 :
        return 9
    else:
        return (n % 9)
     
 
# function to return sum
def PowDigSum(n, x):
    # Find sum of
    # digits in n
    sum = digSum(n)
     
    # Find remainder
    # of exponent
    rem = x % 6
 
    if ((sum == 3 or sum == 6) and x > 1):
            return 9
 
    elif (x == 1):
            return sum
 
    elif (x == 0):
            return 1
 
    elif (rem == 0):
        return digSum(math.pow(sum, 6))
 
    else:
        return digSum(math.pow(sum, rem))
     
# Driver method
n = 33333
x = 332654
print (PowDigSum(n, x))
 
# This code is contributed by Gitanjali

C#




// C# Code for Sum of digit of
// n^x where n and x are very large
using System;
 
class GFG {
 
    // Function to get sum
    // of digits of a number
    static long digSum(long n)
    {
        if (n == 0)
            return 0;
        return (n % 9 == 0) ? 9 : (n % 9);
    }
 
    // Function to return sum
    static long PowDigSum(long n, long x)
    {
        // Find sum of digits in n
        long sum = digSum(n);
 
        // Find remainder of exponent
        long rem = x % 6;
 
        if ((sum == 3 || sum == 6) && x > 1)
            return 9;
 
        else if (x == 1)
            return sum;
 
        else if (x == 0)
            return 1;
 
        else if (rem == 0)
            return digSum((long)Math.Pow(sum, 6));
 
        else
            return digSum((long)Math.Pow(sum, rem));
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 33333;
        int x = 332654;
 
        Console.WriteLine(PowDigSum(n, x));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP Code for Sum of
// digit of n^x where
// function to get sum
// of digits of a number
 
function digSum($n)
    {
        if ($n == 0)
            return 0;
        return ($n % 9 == 0)
                ? 9 : ($n % 9);
    }
 
    // function to return sum
    function PowDigSum($n, $x)
    {
         
        // Find sum of digits in n
        $sum = digSum($n);
 
        // Find remainder of exponent
        $rem = $x % 6;
 
        if (($sum == 3 || $sum == 6)
                        && $x > 1)
            return 9;
 
        else if ($x == 1)
            return $sum;
 
        else if ($x == 0)
            return 1;
 
        else if ($rem == 0)
            return digSum(pow($sum, 6));
 
        else
            return digSum(pow($sum, $rem));
    }
 
    // Driver code
    $n = 33333;
    $x = 332654;
    echo PowDigSum($n, $x);
 
// This code is contributed by aj_36.
?>

Javascript




<script>
 
// JavaScript Code for Sum of
// digit of n^x where
// n and x are very large
  
// function to get sum
// of digits of a number
     function digSum(n)
    {
        if (n == 0)
            return 0;
        return (n % 9 == 0)
                ? 9 : (n % 9);
    }
  
    // function to return sum
     function PowDigSum(n, x)
    {
         
        // Find sum of digits in n
        let sum = digSum(n);
  
        // Find remainder of exponent
        let rem = x % 6;
  
        if ((sum == 3 || sum == 6)
                         && x > 1)
            return 9;
  
        else if (x == 1)
             return sum;
  
        else if (x == 0)
             return 1;
  
        else if (rem == 0)
             return digSum(Math.pow(sum, 6));
  
        else
            return digSum(Math.pow(sum, rem));
    }
  
// Driver code
let n = 33333;
let x = 332654;
document.write(PowDigSum(n, x));
 
</script>

Output: 

9

 

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