# Compute modulus division by a power-of-2-number using Wrapper Class

Prerequisite : Compute modulus division by a power-of-2-number

Now as you know for getting n modulus 2k, we just need to return k bits(from LSB) in binary representation of n. In Java, you can use Wrapper class toBinaryString() method to get binary string representation of a number and getting substring from (str.length()-k) to end. And then by using Integer.parseInt(), you can convert this binary substring to number which is the remainder. Below is the Java program to demonstrate the same.

 `// Java program to Compute modulus ` `// division by a power-of-2-number ` ` `  `class` `Test ` `{ ` `    ``// Driver method ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `num = ``15``; ` `         `  `        ``int` `two_power1 = ``1``; ` `        ``int` `two_power2 = ``2``; ` `        ``int` `two_power3 = ``3``; ` `         `  `        ``String binary = Integer.toBinaryString(num); ` `        ``int` `len = binary.length(); ` `         `  `        ``String rem1 = binary.substring(len-two_power1); ` `        ``String rem2 = binary.substring(len-two_power2); ` `        ``String rem3 = binary.substring(len-two_power3); ` `         `  `        ``int` `reme1 = Integer.parseInt(rem1, ``2``); ` `        ``int` `reme2 = Integer.parseInt(rem2, ``2``); ` `        ``int` `reme3 = Integer.parseInt(rem3, ``2``); ` `         `  `        ``System.out.println(num + ``"%"` `+ ``"2^("` `+ two_power1 + ``") = "` `+ reme1); ` `        ``System.out.println(num + ``"%"` `+ ``"2^("` `+ two_power2 + ``") = "` `+ reme2); ` `        ``System.out.println(num + ``"%"` `+ ``"2^("` `+ two_power3 + ``") = "` `+ reme3); ` `    ``} ` `} `

Output:

```15%2^(1) = 1
15%2^(2) = 3
15%2^(3) = 7
```

This article is contributed by Gaurav Miglani. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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