Check if a number is multiple of 9 using bitwise operators
Given a number n, write a function that returns true if n is divisible by 9, else false. The most simple way to check for n’s divisibility by 9 is to do n%9.
Another method is to sum the digits of n. If sum of digits is multiple of 9, then n is multiple of 9.
The above methods are not bitwise operators based methods and require use of % and /.
The bitwise operators are generally faster than modulo and division operators. Following is a bitwise operator based method to check divisibility by 9.
C++
// C++ program to check if a number // is multiple of 9 using bitwise operators #include <bits/stdc++.h> using namespace std; // Bitwise operator based function to check divisibility by 9 bool isDivBy9( int n) { // Base cases if (n == 0 || n == 9) return true ; if (n < 9) return false ; // If n is greater than 9, then recur for [floor(n/9) - n%8] return isDivBy9(( int )(n >> 3) - ( int )(n & 7)); } // Driver program to test above function int main() { // Let us print all multiples of 9 from 0 to 100 // using above method for ( int i = 0; i < 100; i++) if (isDivBy9(i)) cout << i << " " ; return 0; } |
Java
// Java program to check if a number // is multiple of 9 using bitwise operators import java.lang.*; class GFG { // Bitwise operator based function // to check divisibility by 9 static boolean isDivBy9( int n) { // Base cases if (n == 0 || n == 9 ) return true ; if (n < 9 ) return false ; // If n is greater than 9, then // recur for [floor(n/9) - n%8] return isDivBy9(( int )(n >> 3 ) - ( int )(n & 7 )); } // Driver code public static void main(String arg[]) { // Let us print all multiples of 9 from // 0 to 100 using above method for ( int i = 0 ; i < 100 ; i++) if (isDivBy9(i)) System.out.print(i + " " ); } } // This code is contributed by Anant Agarwal. |
Python3
# Bitwise operator based # function to check divisibility by 9 def isDivBy9(n): # Base cases if (n = = 0 or n = = 9 ): return True if (n < 9 ): return False # If n is greater than 9, # then recur for [floor(n / 9) - n % 8] return isDivBy9(( int )(n>> 3 ) - ( int )(n& 7 )) # Driver code # Let us print all multiples # of 9 from 0 to 100 # using above method for i in range ( 100 ): if (isDivBy9(i)): print (i, " " , end = "") # This code is contributed # by Anant Agarwal. |
C#
// C# program to check if a number // is multiple of 9 using bitwise operators using System; class GFG { // Bitwise operator based function // to check divisibility by 9 static bool isDivBy9( int n) { // Base cases if (n == 0 || n == 9) return true ; if (n < 9) return false ; // If n is greater than 9, then // recur for [floor(n/9) - n%8] return isDivBy9(( int )(n >> 3) - ( int )(n & 7)); } // Driver code public static void Main() { // Let us print all multiples of 9 from // 0 to 100 using above method for ( int i = 0; i < 100; i++) if (isDivBy9(i)) Console.Write(i + " " ); } } // This code is contributed by nitin mittal. |
PHP
<?php // PHP program to check if a number // is multiple of 9 using bitwise // operators // Bitwise operator based function // to check divisibility by 9 function isDivBy9( $n ) { // Base cases if ( $n == 0 || $n == 9) return true; if ( $n < 9) return false; // If n is greater than 9, // then recur for [floor(n/9) - // n%8] return isDivBy9(( $n >> 3) - ( $n & 7)); } // Driver Code // Let us print all multiples // of 9 from 0 to 100 // using above method for ( $i = 0; $i < 100; $i ++) if (isDivBy9( $i )) echo $i , " " ; // This code is contributed by nitin mittal ?> |
Javascript
<script> // javascript program to check if a number // is multiple of 9 using bitwise operators // Bitwise operator based function // to check divisibility by 9 function isDivBy9(n) { // Base cases if (n == 0 || n == 9) return true ; if (n < 9) return false ; // If n is greater than 9, then // recur for [floor(n/9) - n%8] return isDivBy9(parseInt(n >> 3) - parseInt(n & 7)); } // Driver code // Let us print all multiples of 9 from // 0 to 100 using above method for (i = 0; i < 100; i++) if (isDivBy9(i)) document.write(i + " " ); // This code is contributed by Princi Singh </script> |
Output:
0 9 18 27 36 45 54 63 72 81 90 99
Time Complexity: O(log n)
Auxiliary Space: O(logn)
How does this work?
n/9 can be written in terms of n/8 using the following simple formula.
n/9 = n/8 - n/72
Since we need to use bitwise operators, we get the value of floor(n/8) using n>>3 and get value of n%8 using n&7. We need to write above expression in terms of floor(n/8) and n%8.
n/8 is equal to “floor(n/8) + (n%8)/8”. Let us write the above expression in terms of floor(n/8) and n%8
n/9 = floor(n/8) + (n%8)/8 - [floor(n/8) + (n%8)/8]/9 n/9 = floor(n/8) - [floor(n/8) - 9(n%8)/8 + (n%8)/8]/9 n/9 = floor(n/8) - [floor(n/8) - n%8]/9
From above equation, n is a multiple of 9 only if the expression floor(n/8) – [floor(n/8) – n%8]/9 is an integer. This expression can only be an integer if the sub-expression [floor(n/8) – n%8]/9 is an integer. The subexpression can only be an integer if [floor(n/8) – n%8] is a multiple of 9. So the problem reduces to a smaller value which can be written in terms of bitwise operators.
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