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++
#include <bits/stdc++.h>
using namespace std;
bool isDivBy9( int n)
{
if (n == 0 || n == 9)
return true ;
if (n < 9)
return false ;
return isDivBy9(( int )(n >> 3) - ( int )(n & 7));
}
int main()
{
for ( int i = 0; i < 100; i++)
if (isDivBy9(i))
cout << i << " " ;
return 0;
}
|
Java
import java.lang.*;
class GFG {
static boolean isDivBy9( int n)
{
if (n == 0 || n == 9 )
return true ;
if (n < 9 )
return false ;
return isDivBy9(( int )(n >> 3 ) - ( int )(n & 7 ));
}
public static void main(String arg[])
{
for ( int i = 0 ; i < 100 ; i++)
if (isDivBy9(i))
System.out.print(i + " " );
}
}
|
Python3
def isDivBy9(n):
if (n = = 0 or n = = 9 ):
return True
if (n < 9 ):
return False
return isDivBy9(( int )(n>> 3 ) - ( int )(n& 7 ))
for i in range ( 100 ):
if (isDivBy9(i)):
print (i, " " , end = "")
|
C#
using System;
class GFG {
static bool isDivBy9( int n)
{
if (n == 0 || n == 9)
return true ;
if (n < 9)
return false ;
return isDivBy9(( int )(n >> 3) - ( int )(n & 7));
}
public static void Main()
{
for ( int i = 0; i < 100; i++)
if (isDivBy9(i))
Console.Write(i + " " );
}
}
|
PHP
<?php
function isDivBy9( $n )
{
if ( $n == 0 || $n == 9)
return true;
if ( $n < 9)
return false;
return isDivBy9(( $n >> 3) -
( $n & 7));
}
for ( $i = 0; $i < 100; $i ++)
if (isDivBy9( $i ))
echo $i , " " ;
?>
|
Javascript
<script>
function isDivBy9(n)
{
if (n == 0 || n == 9)
return true ;
if (n < 9)
return false ;
return isDivBy9(parseInt(n >> 3) - parseInt(n & 7));
}
for (i = 0; i < 100; i++)
if (isDivBy9(i))
document.write(i + " " );
</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.
Last Updated :
26 May, 2022
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