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

We are given positive integer variables as ‘num’ and ‘x’. The task is to recursively calculate the num ^ x then add the digits of a resultant number till the single digit isn’t achieved and the resultant single digit will be the output.

## Let us see various input output scenarios for this −

**Input **− int num = 2345, int x = 3

**Output **− Recursive sum of digit in n^x, where n and x are very large are: 8

**Explanation **− we are given positive integer values as num and x with the values as 2345 and power as 3. Firstly, calculate 2345 ^ 3 i.e. 12,895,213,625. Now , we will add these digits i.e. 1 + 2 + 8 + 9 + 5 + 2 + 1 + 3 + 6 + 2 + 5 i.e. 44. Now we will add 4 + 4 i.e. 8. Since we have achieved the single digit therefore, output is 8.

**Input **− int num = 3, int x = 3

**Output **− Recursive sum of digit in n^x, where n and x are very large are: 9

**Explanation **− we are given positive integer values as num and x with the values as 3 and power as 3. Firstly, calculate 3 ^ 3 i.e. 9. Since we have achieved the single digit therefore, output is 9 and no further calculation is required.

## Approach used in the below program is as follows

Input an integer variable as num and x and pass the data to the function Recursive_Digit(num, x) for further processing.

Inside the function Recursive_Digit(num, x)

Declare variable ‘total’ as long and set it to call the function total_digits(num) which will return the digit sum of a number passed as an argument.

Declare variable as temp of type long and set it with power % 6

Check IF total = 3 OR total = 6 AND power > 1 then return 9.

ELSE IF, power = 1 then return total.

ELSE IF, power = 0 then return 1.

ELSE IF, temp - 0 then return call to total_digits((long)pow(total, 6))

ELSE, return total_digits((long)pow(total, temp)).

Inside the function long total_digits(long num)

Check IF num = 0 then return 0. Check IF, num % 9 = 0 then return 9.

Else, return num % 9

## Example

#include <bits/stdc++.h> using namespace std; long total_digits(long num){ if(num == 0){ return 0; } if(num % 9 == 0){ return 9; } else{ return num % 9; } } long Recursive_Digit(long num, long power){ long total = total_digits(num); long temp = power % 6; if((total == 3 || total == 6) & power > 1){ return 9; } else if (power == 1){ return total; } else if (power == 0){ return 1; } else if (temp == 0){ return total_digits((long)pow(total, 6)); } else{ return total_digits((long)pow(total, temp)); } } int main(){ int num = 2345; int x = 98754; cout<<"Recursive sum of digit in n^x, where n and x are very large are: "<<Recursive_Digit(num, x); return 0; }

## Output

If we run the above code it will generate the following Output

Recursive sum of digit in n^x, where n and x are very large are: 1

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