# C++ Program Check if a number is multiple of 9 using bitwise operators

Check if a number is multiple of 9 using bitwise operators - Bit Algorithm - Given a number n, write a function that returns true if n is divisible by 9.

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 programming
``````#include<iostream>
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;
}``````

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. 