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

•

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#includeusing namespace std; // Function to find a^b using// binary exponentiationlong 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 equationlong calculate(long X, long N){    int a;     // To store cyclicity    int cyclicity;     // Store cyclicity from 1 - 10    cyclicity = 1;    cyclicity = 4;    cyclicity = 4;    cyclicity = 2;    cyclicity = 1;    cyclicity = 1;    cyclicity = 4;    cyclicity = 4;    cyclicity = 2;    cyclicity = 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;        a = 6;        a = 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 Codeint 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;         // To store cyclicity        int cyclicity[] = new int;         // Store cyclicity from 1 - 10        cyclicity = 1;        cyclicity = 4;        cyclicity = 4;        cyclicity = 2;        cyclicity = 1;        cyclicity = 1;        cyclicity = 4;        cyclicity = 4;        cyclicity = 2;        cyclicity = 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;            a = 6;            a = 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 exponentiationdef 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 equationdef calculate(X, N):     a = 10 *      # To store cyclicity    cyclicity = 11 *      # Store cyclicity from 1 - 10    cyclicity = 1    cyclicity = 4    cyclicity = 4    cyclicity = 2    cyclicity = 1    cyclicity = 1    cyclicity = 4    cyclicity = 4    cyclicity = 2    cyclicity = 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        a = 6        a = 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 Codeif __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 approachusing System;using System.Collections.Generic; class GFG{     // Function to find a^b using// binary exponentiationstatic 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 equationpublic static long calculate(long X,                             long N){    int[] a = new int;     // To store cyclicity    int[] cyclicity = new int;     // Store cyclicity from 1 - 10    cyclicity = 1;    cyclicity = 4;    cyclicity = 4;    cyclicity = 2;    cyclicity = 1;    cyclicity = 1;    cyclicity = 4;    cyclicity = 4;    cyclicity = 2;    cyclicity = 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;        a = 6;        a = 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 codestatic 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

 

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 #include  using namespace std; // This function calculates the last digit of X^N factorialint 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 = 18N = 4print(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 implementationfunction 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)

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