# 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` `?>` |

## Javascript

`<script>` `// javascript 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(parseInt(n >> 3) - parseInt(n & 7));` `}` `// Driver code` ` ` `// Let us prvar all multiples of 9 from` `// 0 to 100 using above method` `for` `(i = 0; i < 100; i++)` ` ` `if` `(isDivBy9(i))` ` ` `document.write(i + ` `" "` `);` `// This code is contributed by Princi Singh` `</script>` |

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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