Sum of Digits in a^n till a single digit

Given two numbers a and n, the task is to find the single sum of digits of a^n (pow(a, n)). In single digit sum, we keep doing sum of digit until a single digit is left.

Examples:

Input : a = 5, n = 4
Output : 4
5^4 = 625 = 6+2+5 = 13
Since 13 has two digits, we
sum again 1 + 3 = 4.

Input : a = 2, n = 8
Output : 4
2^8=256 = 2+5+6 = 13 = 1+3 = 4

A naive approach is to first find a^n, then find sum of digits in a^n using the approach discussed here.

The above approach may cause overflow. A better solution is based on below observation.  

int res = 1;
for (int i=1; i<=n; i++)
{
     res = res*a;
     res = digSum(res);
}

Here digSum() finds single digit sum 
of res. Please refer this for details
of digSum().

Illustration of above pseudo code:  

For example, let a = 5, n = 4. 
After first iteration, 
res = 5
After second iteration, 
res = 7 (Note : 2 + 5 = 7)
After third iteration, 
res = 8 (Note : 3 + 5 = 8)
After 4th iteration, 
res = 4 (Note : 4 + 0 = 4)

We can write a function similar to a fast modular exponentiation to evaluate digSum(a^n) which evaluates this in log(n) steps.
Below is the implementation of above approach:

C++

// CPP program to find single digit
// sum of a^n.
#include <bits/stdc++.h>
using namespace std;

// This function finds single digit
// sum of n.
int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}

// Returns single digit sum of a^n.
// We use modular exponentiation technique.
int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }

    return res;
}

// Driver code
int main()
{
    int a = 9, n = 4;
    cout << powerDigitSum(a, n);
    return 0;
}

Java

// Java program to find single digit
// sum of a^n.

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{


// This function finds single digit
// sum of n.
static int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}

// Returns single digit sum of a^n.
// We use modular exponentiation technique.
static int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n>0) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }

    return res;
}

// Driver code
public static void main(String args[])
{
    int a = 9, n = 4;
    System.out.print(powerDigitSum(a, n));
}
}

Python 3

# Python 3 Program to find single digit
# sum of a^n.

# This function finds single digit
# sum of n.
def digSum(n) :

    if n == 0 :
        return 0

    elif n % 9 == 0 :
        return 9

    else :
        return n % 9

# Returns single digit sum of a^n.
# We use modular exponentiation technique.
def powerDigitSum(a, n) :

    res = 1
    while(n) :

        if n %2 == 1 :
            res = res * digSum(a)
            res = digSum(res)

        a = digSum(digSum(a) * digSum(a))
        n //= 2

    return res


# Driver Code
if __name__ == "__main__" :

    a, n = 9, 4
    print(powerDigitSum(a, n))

# This code is contributed by ANKITRAI1

C#

// C# program to find single
// digit sum of a^n.
class GFG
{

// This function finds single
// digit sum of n.
static int digSum(int n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ?
                      9 : (n % 9);
}

// Returns single digit sum of a^n.
// We use modular exponentiation
// technique.
static int powerDigitSum(int a, int n)
{
    int res = 1;
    while (n > 0)
    {
        if (n % 2 == 1)
        {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }

    return res;
}

// Driver code
static void Main()
{
    int a = 9, n = 4;
    System.Console.WriteLine(powerDigitSum(a, n));
}
}

// This Code is contributed by mits

PHP

<?php
// PHP program to find single
// digit sum of a^n.

// This function finds single
// digit sum of n.
function digSum($n)
{
    if ($n == 0)
    return 0;
    return ($n % 9 == 0) ?
                 9 : ($n % 9);
}

// Returns single digit sum
// of a^n. We use modular
// exponentiation technique.
function powerDigitSum($a, $n)
{
    $res = 1;
    while ($n)
    {
        if ($n % 2 == 1)
        {
            $res = $res * digSum($a);
            $res = digSum($res);
        }
        $a = digSum(digSum($a) *
             digSum($a));
        $n /= 2;
    }

    return $res;
}

// Driver code
$a = 9;
$n = 4;
echo powerDigitSum($a, $n);

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>

// javascript program to find single digit
// sum of a^n.

// This function finds single digit
// sum of n.
function digSum( n)
{
    if (n == 0)
    return 0;
    return (n % 9 == 0) ? 9 : (n % 9);
}

// Returns single digit sum of a^n.
// We use modular exponentiation technique.
function powerDigitSum( a,  n)
{
    let res = 1;
    while (n) {
        if (n % 2 == 1) {
            res = res * digSum(a);
            res = digSum(res);
        }
        a = digSum(digSum(a) * digSum(a));
        n /= 2;
    }

    return res;
}

// Driver code

    let a = 9, n = 4;
    document.write( powerDigitSum(a, n));

// This code is contributed by todaysgaurav


</script>

Output:

9

Time Complexity: O(log₂n)// since in the powerDigitSum function in every call the value of n is divided by 2 until it reaches 1 thus the algorithm takes logarithmic time to execute.

Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant