 Open in App
Not now

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

• Last Updated : 21 Jun, 2017

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.