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
?>


Output:

0 9 18 27 36 45 54 63 72 81 90 99

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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Improved By : nitin mittal

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