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.

#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.

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