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++

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// 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;
}

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Java

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// 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

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Python3

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# 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    

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C#

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// 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

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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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