# 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 ` `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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