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 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 codeint 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 codepublic 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 codefor 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 codepublic static void Main(string[] args){    for (int n = 0; n < 20; n++)    {        Console.Write(lastDigit2PowerN(n) + " ");    }}} // This code is contributed by rutvik_56

Javascript


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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