Last digit in a power of 2

Given a number n, we need to find the last digit of 2ⁿ

Input : n = 4
Output : 6
The last digit in 2^4 = 16 is 6

Input : n = 11
Output : 8
The last digit in 2^11 = 2048 is 8

A Naive Solution is to first compute power = pow(2, n), then find the last digit in power using power % 10. This solution is inefficient and also has integer arithmetic issue for slightly large n.

An Efficient Solution is based on the fact that last digits repeat in cycles of 4 if we leave 2^0 which is 1. Powers of 2 (starting from 2^1) are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, …
We can notice that the last digits are 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, …

1) We compute rem = n % 4. Note that last rem will have value from 0 to 3.
2) We return last digit according to value of remainder.

Remainder   Last Digit
  1            2
  2            4
  3            8
  0            6

Illustration : Let n = 11, rem = n % 4 = 3. Last digit in 2^3 is 8 which is same as last digit of 2^11.

C++

// C++ program to find last digit in a power of 2. 
#include <bits/stdc++.h> 
using namespace std; 
  
int lastDigit2PowerN(int n) 
{ 
  
    // Corner case 
    if (n == 0) 
        return 1; 
  
    // Find the shift in current cycle 
    // and return value accordingly 
    else if (n % 4 == 1) 
        return 2; 
    else if (n % 4 == 2) 
        return 4; 
    else if (n % 4 == 3) 
        return 8; 
    else
        return 6; // When n % 4 == 0 
} 
  
// Driver code 
int main() 
{ 
    for (int n = 0; n < 20; n++) 
        cout << lastDigit2PowerN(n) << " "; 
    return 0; 
}

Java

// Java program to find last  
// digit in a power of 2. 
import java.io.*;  
import java.util.*;  
  
class GFG{ 
      
static int lastDigit2PowerN(int n) 
{ 
      
    // Corner case 
    if (n == 0) 
        return 1; 
  
    // Find the shift in current cycle 
    // and return value accordingly 
    else if (n % 4 == 1) 
        return 2; 
    else if (n % 4 == 2) 
        return 4; 
    else if (n % 4 == 3) 
        return 8; 
    else
        return 6; // When n % 4 == 0 
} 
  
// Driver code  
public static void main(String[] args)  
{  
    for (int n = 0; n < 20; n++) 
    System.out.print(lastDigit2PowerN(n) + " "); 
}  
} 
  
// This code is contributed by coder001 

Python3

# Python3 program to find last  
# digit in a power of 2. 
def lastDigit2PowerN(n): 
      
    # Corner case  
    if n == 0:  
        return 1
  
    # Find the shift in current cycle  
    # and return value accordingly  
    elif n % 4 == 1:  
        return 2
    elif n % 4 == 2:  
        return 4
    elif n % 4 == 3: 
        return 8
    else: 
        return 6 # When n % 4 == 0  
  
# Driver code  
for n in range(20): 
    print(lastDigit2PowerN(n), end = " ") 
  
# This code is contributed by divyeshrabadiya07     

C#

// C# program to find last  
// digit in a power of 2. 
using System; 
class GFG{ 
      
static int lastDigit2PowerN(int n) 
{ 
      
    // Corner case 
    if (n == 0) 
        return 1; 
  
    // Find the shift in current cycle 
    // and return value accordingly 
    else if (n % 4 == 1) 
        return 2; 
    else if (n % 4 == 2) 
        return 4; 
    else if (n % 4 == 3) 
        return 8; 
    else
        return 6; // When n % 4 == 0 
} 
  
// Driver code  
public static void Main(string[] args)  
{  
    for (int n = 0; n < 20; n++) 
    { 
        Console.Write(lastDigit2PowerN(n) + " "); 
    } 
}  
} 
  
// This code is contributed by rutvik_56 

Output:

1 2 4 8 6 2 4 8 6 2 4 8 6 2 4 8 6 2 4 8

Time Complexity: O(1)
Auxiliary Space: O(1)

Can we generalize it for any input numbers? Please refer Find Last Digit of a^b for Large Numbers

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My Personal Notes

C++

Java

Python3

C#

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