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Last digit in a power of 2

  • Last Updated : 27 Apr, 2021

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 an integer arithmetic issue for slightly large n.
An Efficient Solution is based on the fact that the 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 the last rem will have a value from 0 to 3. 
2) We return the last digit according to the value of the 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

Javascript




<script>
 
      // JavaScript program to find
      // last digit in a power of 2.
 
      function lastDigit2PowerN(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
      for (var n = 0; n < 20; n++)
      document.write(lastDigit2PowerN(n) + " ");
       
    </script>
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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