# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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  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;         // 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 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;             // 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);      }  }

Output:

6


Time Complexity: O(1)

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