# Check if n is divisible by power of 2 without using arithmetic operators

Given two positive integers n and m. The problem is to check whether n is divisible by 2m or not without using arithmetic operators.

Examples:

```Input : n = 8, m = 2
Output : Yes

Input : n = 14, m = 3
Output : No```

Approach: If a number is divisible by 2 then it has its least significant bit (LSB) set to 0, if divisible by 4 then two LSB’s set to 0, if by 8 then three LSB’s set to 0, and so on. Keeping this in mind, a number n is divisible by 2m if (n & ((1 << m) – 1)) is equal to 0 else not.

## C++

 `// C++ implementation to check whether n` `// is divisible by pow(2, m)` `#include `   `using` `namespace` `std;`   `// function to check whether n` `// is divisible by pow(2, m)` `bool` `isDivBy2PowerM(unsigned ``int` `n,` `                    ``unsigned ``int` `m)` `{` `    ``// if expression results to 0, then` `    ``// n is divisible by pow(2, m)` `    ``if` `((n & ((1 << m) - 1)) == 0)` `        ``return` `true``;`   `    ``// n is not divisible` `    ``return` `false``;` `}`   `// Driver program to test above` `int` `main()` `{` `    ``unsigned ``int` `n = 8, m = 2;` `    ``if` `(isDivBy2PowerM(n, m))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `// JAVA Code for Check if n is divisible  ` `// by power of 2 without using arithmetic ` `// operators` `import` `java.util.*;`   `class` `GFG {` `    `  `    ``// function to check whether n` `    ``// is divisible by pow(2, m)` `    ``static` `boolean` `isDivBy2PowerM(``int` `n,` `                                    ``int` `m)` `    ``{` `        ``// if expression results to 0, then` `        ``// n is divisible by pow(2, m)` `        ``if` `((n & ((``1` `<< m) - ``1``)) == ``0``)` `            ``return` `true``;` `     `  `        ``// n is not divisible` `        ``return` `false``;` `    ``}` `    `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `            ``int` `n = ``8``, m = ``2``;` `            `  `            ``if` `(isDivBy2PowerM(n, m))` `                ``System.out.println(``"Yes"``);` `            ``else` `                ``System.out.println(``"No"``);` `    ``}` `}`   `// This code is contributed by Arnav Kr. Mandal.`

## Python3

 `# Python3 implementation to check ` `# whether n is divisible by pow(2, m)`   `# function to check whether n` `# is divisible by pow(2, m)` `def` `isDivBy2PowerM (n, m):` `    `  `    ``# if expression results to 0, then` `    ``# n is divisible by pow(2, m)` `    ``if` `(n & ((``1` `<< m) ``-` `1``)) ``=``=` `0``:` `        ``return` `True` `        `  `    ``# n is not divisible` `    ``return` `False`   `# Driver program to test above` `n ``=` `8` `m ``=` `2` `if` `isDivBy2PowerM(n, m):` `    ``print``(``"Yes"``)` `else``:` `    ``print``( ``"No"``)` `    `  `# This code is contributed by "Sharad_Bhardwaj".`

## C#

 `// C# Code for Check if n is divisible` `// by power of 2 without using arithmetic` `// operators` `using` `System;`   `class` `GFG {`   `    ``// function to check whether n` `    ``// is divisible by pow(2, m)` `    ``static` `bool` `isDivBy2PowerM(``int` `n, ``int` `m)` `    ``{` `        ``// if expression results to 0, then` `        ``// n is divisible by pow(2, m)` `        ``if` `((n & ((1 << m) - 1)) == 0)` `            ``return` `true``;`   `        ``// n is not divisible` `        ``return` `false``;` `    ``}`   `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 8, m = 2;`   `        ``if` `(isDivBy2PowerM(n, m))` `            ``Console.Write(``"Yes"``);` `        ``else` `            ``Console.Write(``"No"``);` `    ``}` `}`   `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

Time Complexity : O(1)

Auxiliary Space: O(1)

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