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

• Last Updated : 16 Nov, 2021

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 â‰¥ 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  #include 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

 // Java program for the above approach 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

 

Output:

6

Time Complexity: O(1)

Auxiliary Space: O(11)

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