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 6Input : 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 |
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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