# Round to next smaller multiple of 8

Given an unsigned integer x. Round it down to the next smaller multiple of 8 using bitwise operations only.**Examples:**

Input : 35 Output : 32 Input : 40 Output : 40 As 40 is already a multiple of 8. So, no modification is done.

**Solution 1:** A naive approach to solve this problem using arithmetic operators is :

Let x be the number then,

x = x – (x % 8)

This will round down x to the next smaller multiple of 8. But we are not allowed to use arithmetic operators.**Solution 2:** An efficient approach to solve this problem using bitwise AND operation is: x = x & (-8)

This will round down x to the next smaller multiple of 8. The idea is based on the fact that last three bits in a multiple of 8 must be 0,

Below is the implementation of above idea:

## C++

`// CPP program to find next smaller` `// multiple of 8.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `RoundDown(` `int` `& a)` `{` ` ` `return` `a & (-8);` `}` `int` `main()` `{` ` ` `int` `x = 39;` ` ` `cout << RoundDown(x);` ` ` `return` `0;` `}` |

## Java

`//Java program to find next smaller` `// multiple of 8.` `import` `java.io.*;` `class` `GFG {` `static` `int` `RoundDown(` `int` `a)` `{` ` ` `return` `a & (-` `8` `);` `}` ` ` `public` `static` `void` `main (String[] args) {` ` ` `int` `x = ` `39` `;` ` ` `System.out.println (RoundDown(x));` ` ` `}` `}` `//This Code is Contributed by ajit` |

## Python3

`# Python 3 program to find next` `# smaller multiple of 8.` `def` `RoundDown(a):` ` ` `return` `a & (` `-` `8` `)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `x ` `=` `39` ` ` `print` `(RoundDown(x))` `# This code is contributed` `# by Surendra_Gangwar` |

## C#

`// C# program to find next smaller` `// multiple of 8.` `using` `System;` `class` `GFG` `{` `static` `int` `RoundDown(` `int` `a)` `{` ` ` `return` `a & (-8);` `}` `public` `static` `void` `Main()` `{` ` ` `int` `x = 39;` ` ` `Console.Write(RoundDown(x));` `}` `}` `// This code is contributed` `// by Akanksha Rai` |

## PHP

`<?php` `// PHP program to find next smaller` `// multiple of 8.` `function` `RoundDown(` `$a` `)` `{` ` ` `return` `(` `$a` `& (-8));` `}` `// Driver Code` `$x` `= 39;` `echo` `RoundDown(` `$x` `);` `// This code is contributed by jit_t` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to find next smaller multiple of 8.` ` ` ` ` `function` `RoundDown(a)` ` ` `{` ` ` `return` `a & (-8);` ` ` `}` ` ` ` ` `let x = 39;` ` ` `document.write(RoundDown(x));` `</script>` |

**Output:**

32

**Time Complexity:** The time complexity of this approach is O(1) **Space Complexity:** The space complexity of this approach is O(1)

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