Last digit of a number raised to last digit of N factorial

Given two number X and N, the task is to find the last digit of X raised to last digit of N factorial, i.e. .
Examples:

Input: X = 5, N = 2 
Output: 5 
Explanation:
Since, 2! mod 10 = 2 
therefore 5² = 25 and the last digit of 25 is 5.
Input: X = 10, N = 4 
Output: 0 
Explanation:
Since, 4! mod 10 = 24 mod 10 = 4 
therefore 10⁴ = 10000 and the last digit of 10000 is 0. 

Approach: The most efficient way to solve this problem is to find any pattern in the required last digit, with the help of last digit of N! and last digit of X raised to Y
Below is the various observation of the above-given equation: 

• If N = 0 or N = 1, then the last digit is 1 or respectively.
• Since 5! is 120, therefore for N ? 5 the value of (N! mod 10) will be zero.
• Now we are left with digit 2, 3, 4. For this we have:  

for N = 2, 
N! mod 10 = 2! mod 10 = 2
for N = 3, 
N! mod 10 = 3! mod 10 = 6
for N = 4, 
N! mod 10 = 4! mod 10 = 24 mod 10 = 4
Now for X², X⁴, and X⁶
we will check that after which nth power of Xⁿ the value of last digit repeats, 
i.e, after which nth power of last digit of Xⁿ the value of last digit repeats. 

• Below is the table for what power of the last digit from 0 to 9 in any number repeats: 
  

Number	Cyclicity
0	1
1	1
2	4
3	4
4	2
5	1
6	1
7	4
8	4
9	2

• 
  

Below are the steps based on the above observations:  

1. If X is not a multiple of 10 then divide the evaluated value of by cyclicity of the last digit of X. If remainder(say r) is 0 then do the following: 
   
   • If the last digit of X is any of 2, 4, 6, or 8 then the answer will be 6.
   • If the last digit of X is any of 1, 3, 7, or 9 then the answer will be 1.
   • If the last digit of X is 5 then answer will be 5.
2. Else if remainder(say r) is a non-zero then answer is , where ‘l’ is the last digit of X.
3. Else if X is a multiple of 10 then the answer will be 0 always.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find a^b using
// binary exponentiation
long power(long a, long b, long c)
{

    // Initialise result
    long result = 1;

    while (b > 0)
    {

        // If b is odd then,
        // multiply result by a
        if ((b & 1) == 1)
        {
            result = (result * a) % c;
        }

        // b must be even now
        // Change b to b/2
        b /= 2;

        // Change a = a^2
        a = (a * a) % c;
    }
    return result;
}

// Function to find the last digit
// of the given equation
long calculate(long X, long N)
{
    int a[10];

    // To store cyclicity
    int cyclicity[11];

    // Store cyclicity from 1 - 10
    cyclicity[1] = 1;
    cyclicity[2] = 4;
    cyclicity[3] = 4;
    cyclicity[4] = 2;
    cyclicity[5] = 1;
    cyclicity[6] = 1;
    cyclicity[7] = 4;
    cyclicity[8] = 4;
    cyclicity[9] = 2;
    cyclicity[10] = 1;

    // Observation 1
    if (N == 0 || N == 1)
    {
        return (X % 10);
    }

    // Observation 3
    else if (N == 2 || N == 3 || N == 4)
    {
        long temp = (long)1e18;

        // To store the last digits
        // of factorial 2, 3, and 4
        a[2] = 2;
        a[3] = 6;
        a[4] = 4;

        // Find the last digit of X
        long v = X % 10;

        // Step 1
        if (v != 0)
        {
            int u = cyclicity[(int)v];

            // Divide a[N] by cyclicity
            // of v
            int r = a[(int)N] % u;

            // If remainder is 0
            if (r == 0)
            {

                // Step 1.1
                if (v == 2 || v == 4 ||
                    v == 6 || v == 8)
                {
                    return 6;
                }

                // Step 1.2
                else if (v == 5)
                {
                    return 5;
                }

                // Step 1.3
                else if (v == 1 || v == 3 ||
                         v == 7 || v == 9)
                {
                    return 1;
                }
            }

            // If r is non-zero,
            // then return (l^r) % 10
            else
            {
                return (power(v, r, temp) % 10);
            }
        }

        // Else return 0
        else
        {
            return 0;
        }
    }

    // Else return 1
    return 1;
}

// Driver Code
int main()
{

    // Given Numbers
    int X = 18;
    int N = 4;

    // Function Call
    long result = calculate(X, N);

    // Print the result
    cout << result;
}

// This code is contributed by spp____

Java

import java.util.*;

class TestClass {

    // Function to find a^b using
    // binary exponentiation
    public static long power(long a, long b, long c) {
        // Initialise result
        long result = 1;

        while (b > 0) {
            // If b is odd then,
            // multiply result by a
            if ((b & 1) == 1) {
                result = (result * a) % c;
            }

            // b must be even now
            // Change b to b/2
            b /= 2;

            // Change a = a^2
            a = (a * a) % c;
        }
        return result;
    }

    // Function to find the last digit
    // of the given equation
    public static long calculate(long X, long N) {
        int a[] = new int[10];

        // To store cyclicity
        int cyclicity[] = new int[11];

        // Store cyclicity from 1 - 10
        cyclicity[1] = 1;
        cyclicity[2] = 4;
        cyclicity[3] = 4;
        cyclicity[4] = 2;
        cyclicity[5] = 1;
        cyclicity[6] = 1;
        cyclicity[7] = 4;
        cyclicity[8] = 4;
        cyclicity[9] = 2;
        cyclicity[10] = 1;

        // Observation 1
        if (N == 0 || N == 1) {
            return (X % 10);
        }
        // Observation 3
        else if (N == 2 || N == 3 || N == 4) {
            long temp = (long) 1e18;

            // To store the last digits
            // of factorial 2, 3, and 4
            a[2] = 2;
            a[3] = 6;
            a[4] = 4;

            // Find the last digit of X
            long v = X % 10;

            // Step 1
            if (v != 0) {
                int u = cyclicity[(int) v];

                // Divide a[N] by cyclicity
                // of v
                int r = a[(int) N] % u;

                // If remainder is 0
                if (r == 0) {

                    // Step 1.1
                    if (v == 2 || v == 4 || v == 6 || v == 8) {
                        return 6;
                    }

                    // Step 1.2
                    else if (v == 5) {
                        return 5;
                    }

                    // Step 1.3
                    else if (v == 1 || v == 3 || v == 7 || v == 9) {
                        return 1;
                    }
                }

                // If r is non-zero,
                // then return (l^r) % 10
                else {
                    return (power(v, r, temp) % 10);
                }
            }

            // Else return 0
            else {
                return 0;
            }
        }

        // Else return 1
        return 1;
    }

    // Driver's Code
    public static void main(String args[])
        throws Exception
    {

        // Given Numbers
        int X = 18;
        int N = 4;

        // Function Call
        long result = calculate(X, N);

        // Print the result
        System.out.println(result);
    }
}

Python3

# Python3 program for the above approach

# Function to find a^b using
# binary exponentiation
def power(a, b, c):

    # Initialise result
    result = 1

    while (b > 0):

        # If b is odd then,
        # multiply result by a
        if ((b & 1) == 1):
            result = (result * a) % c

        # b must be even now
        # Change b to b/2
        b //= 2

        # Change a = a^2
        a = (a * a) % c

    return result

# Function to find the last digit
# of the given equation
def calculate(X, N):

    a = 10 * [0]

    # To store cyclicity
    cyclicity = 11 * [0]

    # Store cyclicity from 1 - 10
    cyclicity[1] = 1
    cyclicity[2] = 4
    cyclicity[3] = 4
    cyclicity[4] = 2
    cyclicity[5] = 1
    cyclicity[6] = 1
    cyclicity[7] = 4
    cyclicity[8] = 4
    cyclicity[9] = 2
    cyclicity[10] = 1

    # Observation 1
    if (N == 0 or N == 1):
        return (X % 10)

    # Observation 3
    elif (N == 2 or N == 3 or N == 4):
        temp = 1e18;

        # To store the last digits
        # of factorial 2, 3, and 4
        a[2] = 2
        a[3] = 6
        a[4] = 4

        # Find the last digit of X
        v = X % 10

        # Step 1
        if (v != 0):
            u = cyclicity[v]

            # Divide a[N] by cyclicity
            # of v
            r = a[N] % u

            # If remainder is 0
            if (r == 0):

                # Step 1.1
                if (v == 2 or v == 4 or
                    v == 6 or v == 8):
                    return 6

                # Step 1.2
                elif (v == 5):
                    return 5

                # Step 1.3
                elif (v == 1 or v == 3 or
                      v == 7 or v == 9):
                    return 1

            # If r is non-zero,
            # then return (l^r) % 10
            else:
                return (power(v, r, temp) % 10)

        # Else return 0
        else:
            return 0

    # Else return 1
    return 1

# Driver Code
if __name__ == "__main__":

    # Given numbers
    X = 18
    N = 4

    # Function call
    result = calculate(X, N)

    # Print the result
    print(result)

# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find a^b using
// binary exponentiation
static long power(long a, long b, long c)
{

    // Initialise result
    long result = 1;

    while (b > 0)
    {

        // If b is odd then,
        // multiply result by a
        if ((b & 1) == 1)
        {
            result = (result * a) % c;
        }

        // b must be even now
        // Change b to b/2
        b /= 2;

        // Change a = a^2
        a = (a * a) % c;
    }
    return result;
}

// Function to find the last digit
// of the given equation
public static long calculate(long X,
                             long N)
{
    int[] a = new int[10];

    // To store cyclicity
    int[] cyclicity = new int[11];

    // Store cyclicity from 1 - 10
    cyclicity[1] = 1;
    cyclicity[2] = 4;
    cyclicity[3] = 4;
    cyclicity[4] = 2;
    cyclicity[5] = 1;
    cyclicity[6] = 1;
    cyclicity[7] = 4;
    cyclicity[8] = 4;
    cyclicity[9] = 2;
    cyclicity[10] = 1;

    // Observation 1
    if (N == 0 || N == 1)
    {
        return (X % 10);
    }
    // Observation 3
    else if (N == 2 || N == 3 || N == 4)
    {
        long temp = (long)1e18;

        // To store the last digits
        // of factorial 2, 3, and 4
        a[2] = 2;
        a[3] = 6;
        a[4] = 4;

        // Find the last digit of X
        long v = X % 10;

        // Step 1
        if (v != 0)
        {
            int u = cyclicity[(int)v];

            // Divide a[N] by cyclicity
            // of v
            int r = a[(int)N] % u;

            // If remainder is 0
            if (r == 0)
            {

                // Step 1.1
                if (v == 2 || v == 4 ||
                    v == 6 || v == 8)
                {
                    return 6;
                }

                // Step 1.2
                else if (v == 5)
                {
                    return 5;
                }

                // Step 1.3
                else if ( v == 1 || v == 3 ||
                          v == 7 || v == 9)
                {
                    return 1;
                }
            }

            // If r is non-zero,
            // then return (l^r) % 10
            else
            {
                return (power(v, r, temp) % 10);
            }
        }

        // Else return 0
        else
        {
            return 0;
        }
    }

    // Else return 1
    return 1;
}

// Driver code
static void Main()
{

    // Given numbers
    int X = 18;
    int N = 4;

    // Function call
    long result = calculate(X, N);

    // Print the result
    Console.Write(result);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>
      // JavaScript program for the above approach

      // Function to find a^b using
      // binary exponentiation
      function power(a, b, c) {
        // Initialise result
        var result = 1;

        while (b > 0) {
          // If b is odd then,
          // multiply result by a
          if ((b & 1) == 1) {
            result = (result * a) % c;
          }

          // b must be even now
          // Change b to b/2
          b /= 2;

          // Change a = a^2
          a = (a * a) % c;
        }
        return result;
      }

      // Function to find the last digit
      // of the given equation
      function calculate(X, N) {
        var a = [...Array(10)];

        // To store cyclicity
        var cyclicity = [...Array(11)];

        // Store cyclicity from 1 - 10
        cyclicity[1] = 1;
        cyclicity[2] = 4;
        cyclicity[3] = 4;
        cyclicity[4] = 2;
        cyclicity[5] = 1;
        cyclicity[6] = 1;
        cyclicity[7] = 4;
        cyclicity[8] = 4;
        cyclicity[9] = 2;
        cyclicity[10] = 1;

        // Observation 1
        if (N == 0 || N == 1) {
          return X % 10;
        }

        // Observation 3
        else if (N == 2 || N == 3 || N == 4) {
          var temp = 1e18;

          // To store the last digits
          // of factorial 2, 3, and 4
          a[2] = 2;
          a[3] = 6;
          a[4] = 4;

          // Find the last digit of X
          var v = X % 10;

          // Step 1
          if (v != 0) {
            var u = cyclicity[parseInt(v)];

            // Divide a[N] by cyclicity
            // of v
            var r = a[parseInt(N)] % u;

            // If remainder is 0
            if (r == 0)
            {

              // Step 1.1
              if (v == 2 || v == 4 || v == 6 || v == 8) {
                return 6;
              }

              // Step 1.2
              else if (v == 5) {
                return 5;
              }

              // Step 1.3
              else if (v == 1 || v == 3 || v == 7 || v == 9) {
                return 1;
              }
            }

            // If r is non-zero,
            // then return (l^r) % 10
            else {
              return power(v, r, temp) % 10;
            }
          }

          // Else return 0
          else {
            return 0;
          }
        }

        // Else return 1
        return 1;
      }

      // Driver Code
      // Given Numbers
      var X = 18;
      var N = 4;

      // Function Call
      var result = calculate(X, N);

      // Print the result
      document.write(result);

      // This code is contributed by rdtank.
    </script>

Output:

6

Time Complexity: O(logn) because it is using a while loop
Auxiliary Space: O(11)

Approach#2: using for loop

We can first calculate the last digit of N factorial and then take the last digit of X raised to this last digit. We can use modular arithmetic to calculate the last digit of N factorial and to take the last digit of X raised to this value.

Algorithm

1. Calculate the last digit of N factorial using modular arithmetic.
2. Calculate the last digit of X raised to this last digit using modular arithmetic.
3. Return the result.

C++

#include <cmath>
#include <iostream>

using namespace std;

// This function calculates the last digit of X^N factorial
int lastDigit(int X, int N)
{
    // Initialize the last digit of N factorial as 1
    int lastDigitNFactorial = 1;

    // Calculate the last digit of N factorial
    for (int i = 2; i <= N; i++) {
        lastDigitNFactorial
            = (lastDigitNFactorial * (i % 10)) % 10;
    }

    // Calculate the last digit of X^(N factorial)
    int lastDigitXToLastDigitNFactorial
        = (int)pow(X % 10, lastDigitNFactorial) % 10;

    // Return the result
    return lastDigitXToLastDigitNFactorial;
}

int main()
{
    // Initialize X and N
    int X = 18;
    int N = 4;

    // Print the result of lastDigit(X, N)
    cout << lastDigit(X, N) << endl;

    return 0;
}

Java

public class Main {
    public static int lastDigit(int X, int N)
    {
        int lastDigitNFactorial = 1;
        for (int i = 2; i <= N; i++) {
            lastDigitNFactorial
                = (lastDigitNFactorial * (i % 10)) % 10;
        }
        int lastDigitXToLastDigitNFactorial
            = (int)Math.pow(X % 10, lastDigitNFactorial)
              % 10;
        return lastDigitXToLastDigitNFactorial;
    }

    public static void main(String[] args)
    {
        int X = 18;
        int N = 4;
        System.out.println(lastDigit(X, N));
    }
}

Python3

def last_digit(X, N):
    last_digit_n_factorial = 1
    for i in range(2, N+1):
        last_digit_n_factorial = (last_digit_n_factorial * (i % 10)) % 10
    last_digit_X_to_last_digit_n_factorial = pow(
        X % 10, last_digit_n_factorial, 10)
    return last_digit_X_to_last_digit_n_factorial


X = 18
N = 4
print(last_digit(X, N))

C#

using System;

class Program {


    // This function calculates the last digit of X^N
    // factorial
    static int lastDigit(int X, int N)
    {
        // Initialize the last digit of N factorial as 1
        int lastDigitNFactorial = 1;

        // Calculate the last digit of N factorial
        for (int i = 2; i <= N; i++) {
            lastDigitNFactorial
                = (lastDigitNFactorial * (i % 10)) % 10;
        }

        // Calculate the last digit of X^(N factorial)
        int lastDigitXToLastDigitNFactorial
            = (int)Math.Pow(X % 10, lastDigitNFactorial)
              % 10;

        // Return the result
        return lastDigitXToLastDigitNFactorial;
    }
  static void Main(string[] args)
    {
        // Initialize X and N
        int X = 18;
        int N = 4;

        // Print the result of lastDigit(X, N)
        Console.WriteLine(lastDigit(X, N));
    }
}

Javascript

// Javascript program implementation
function last_digit(X, N) {
    let last_digit_n_factorial = 1;
    for (let i = 2; i <= N; i++) {
        last_digit_n_factorial = (last_digit_n_factorial * (i % 10)) % 10;
    }
    let last_digit_X_to_last_digit_n_factorial = Math.pow(X % 10, last_digit_n_factorial) % 10;
    return last_digit_X_to_last_digit_n_factorial;
}

let X = 18;
let N = 4;
console.log(last_digit(X, N));

Output

6

Time complexity: O(N), where N is the input
Space complexity: O(1)