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Check if a number is multiple of 9 using bitwise operators
• Difficulty Level : Hard
• Last Updated : 05 Apr, 2021

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 ``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

 `> 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

 ``

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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