Given a number n, check if it is divisible by 17 using bitwise operators.
Examples:
Input : n = 34 Output : 34 is divisible by 17 Input : n = 43 Output : 43 is not divisible by 17
A naive approach will be to check it by % operator if it leaves a remainder of 0.
To do division using Bitwise operators, we must rewrite the expression in powers of 2.
n/17 = (16*n)/(17*16)
= (17 - 1)*n/(17*16)
= (n/16) - (n/(17*16))
We can rewrite n/16 as floor(n/16) + (n%16)/16 using general rule of division.
n/17 = floor(n/16) + (n%16)/16 -
(floor(n/16) + (n%16)/16)/17
= floor(n/16) - (floor(n/16) -
17*(n%16)/16 + (n%16)/16)/17
= floor(n/16) - (floor(n/16)-n%16)/17
The left-hand-side of this equation is n/17. That will be an integer only when the right-hand-side is an integer. floor(n/16) is an integer by definition. So the whole left-hand-side would be an integer if (floor(n/16)-n%16)/17 is also an integer.
This implies n is divisible by 17 if (floor(n/16)-n%16) is divisible by 17.
(floor(n/16)-n%16) can be written in bitwise as (int)(n>>4) – (int)(n&15) where n>>4 means n/16 and n&15 means n%16
Below is the implementation of the above approach:
CPP
// CPP program to check if a number is// divisible by 17 or not using bitwise// operator.#include <bits/stdc++.h>using namespace std;// function to check recursively if the// number is divisible by 17 or notbool isDivisibleby17(int n){ // if n=0 or n=17 then yes if (n == 0 || n == 17) return true; // if n is less than 17, not // divisible by 17 if (n < 17) return false; // reducing the number by floor(n/16) // - n%16 return isDivisibleby17((int)(n >> 4) - (int)(n & 15));}// driver code to check the above functionint main(){ int n = 35; if (isDivisibleby17(n)) cout << n << " is divisible by 17"; else cout << n << " is not divisible by 17"; return 0;} |
Java
// Java program to check if a number is// divisible by 17 or not using bitwise// operator.class GFG{ // function to check recursively if the // number is divisible by 17 or not static boolean isDivisibleby17(int n) { // if n=0 or n=17 then yes if (n == 0 || n == 17) return true; // if n is less than 17, not // divisible by 17 if (n < 17) return false; // reducing the number by // floor(n/16) - n%16 return isDivisibleby17((int)(n >> 4) - (int)(n & 15)); } // driver function public static void main(String[] args) { int n = 35; if (isDivisibleby17(n) == true) System.out.printf ("%d is divisible by 17",n); else System.out.printf ("%d is not divisible by 17",n); }}// This code is contributed by// Smitha Dinesh Semwal |
Python3
# Python 3 program to# check if a number is# divisible by 17 or# not using bitwise# operator.# function to check recursively if the# number is divisible by 17 or notdef isDivisibleby17(n): # if n=0 or n=17 then yes if (n == 0 or n == 17): return True # if n is less than 17, not # divisible by 17 if (n < 17): return False # reducing the number by floor(n/16) # - n%16 return isDivisibleby17((int)(n >> 4) - (int)(n & 15))# driver code to check the above functionn = 35if (isDivisibleby17(n)): print(n,"is divisible by 17")else: print(n,"is not divisible by 17")# This code is contributed by# Smitha Dinesh Semwal |
C#
// C# program to check if a number is// divisible by 17 or not using bitwise// operator.using System;class GFG{ // function to check recursively if the // number is divisible by 17 or not static bool isDivisibleby17(int n) { // if n=0 or n=17 then yes if (n == 0 || n == 17) return true; // if n is less than 17, not // divisible by 17 if (n < 17) return false; // reducing the number by // floor(n/16) - n%16 return isDivisibleby17((int)(n >> 4) - (int)(n & 15)); } // Driver function public static void Main() { int n = 35; if (isDivisibleby17(n) == true) Console.WriteLine (n +"is divisible by 17"); else Console.WriteLine ( n+ " is not divisible by 17"); }}// This code is contributed by// vt_m |
PHP
<?php// php program to check if a // number is divisible by 17// or not using bitwise// operator.// function to check recursively // if the number is divisible // by 17 or notfunction isDivisibleby17($n){ // if n=0 or n=17 then yes if ($n == 0 || $n == 17) return true; // if n is less than 17, not // divisible by 17 if ($n < 17) return false; // reducing the number by floor(n/16) // - n%16 return isDivisibleby17((int)($n >> 4) - (int)($n & 15));} // Driver Code $n = 35; if (isDivisibleby17($n)) echo $n." is divisible by 17"; else echo $n." is not divisible by 17";// This code is contributed by mits ?> |
Javascript
<script>// JavaScript program to check if a number is // divisible by 17 or not using bitwise // operator. // function to check recursively if the // number is divisible by 17 or not function isDivisibleby17(n) { // if n=0 or n=17 then yes if (n == 0 || n == 17) return true; // if n is less than 17, not // divisible by 17 if (n < 17) return false; // reducing the number by floor(n/16) // - n%16 return isDivisibleby17(Math.floor(n >> 4) - Math.floor(n & 15)); } // driver code to check the above function let n = 35; if (isDivisibleby17(n)) document.write(n + " is divisible by 17"); else document.write(n + " is not divisible by 17"); // This code is contributed by Surbhi Tyagi.</script> |
Output:
35 is not divisible by 17
Time Complexity: O(log16N), as we are using recursion and in each call we are decrementing by division of 16.
Auxiliary Space: O(1), as we are not using any extra space.
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