# Count numbers which are divisible by all the numbers from 2 to 10

Given an integer **N**, the task is to find the count of numbers from **1** to **N** which are divisible by all the numbers from **2** to **10**.

**Examples:**

Input:N = 3000Output:1

2520 is the only number below 3000 which is divisible by all the numbers from 2 to 10.

Input:N = 2000Output:0

**Approach:** Let’s factorize numbers from 2 to 10.

2 = 2

3 = 3

4 = 2^{2}

5 = 5

6 = 2 * 3

7 = 7

8 = 2^{3}

9 = 3^{2}

10 = 2 * 5

If a number is divisible by all the numbers from **2** to **10**, its factorization should contain **2** at least in the power of **3**, **3** at least in the power of **2**, **5** and **7** at least in the power of **1**. So it can be written as:

x * 2^{3}* 3^{2}* 5 * 7 i.e. x * 2520.

So any number divisible by **2520** is divisible by all the numbers from **2** to **10**. So, the count of such numbers is **N / 2520**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the count of numbers` `// from 1 to n which are divisible by` `// all the numbers from 2 to 10` `int` `countNumbers(` `int` `n)` `{` ` ` `return` `(n / 2520);` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3000;` ` ` `cout << countNumbers(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` `// Function to return the count of numbers` `// from 1 to n which are divisible by` `// all the numbers from 2 to 10` `static` `int` `countNumbers(` `int` `n)` `{` ` ` `return` `(n / ` `2520` `);` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `n = ` `3000` `;` ` ` `System.out.println(countNumbers(n));` `}` `}` `// This code is contributed by Arnab Kundu` |

## Python3

`# Python3 implementation of the approach` `# Function to return the count of numbers` `# from 1 to n which are divisible by` `# all the numbers from 2 to 10` `def` `countNumbers(n):` ` ` `return` `n ` `/` `/` `2520` `# Driver code` `n ` `=` `3000` `print` `(countNumbers(n))` `# This code is contributed` `# by Shrikant13` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` `// Function to return the count of numbers` `// from 1 to n which are divisible by` `// all the numbers from 2 to 10` `static` `int` `countNumbers(` `int` `n)` `{` ` ` `return` `(n / 2520);` `}` `// Driver code` `public` `static` `void` `Main(String []args)` `{` ` ` `int` `n = 3000;` ` ` `Console.WriteLine(countNumbers(n));` `}` `}` `// This code is contributed by Arnab Kundu` |

## PHP

`<?php` `// PHP implementation of the approach` `// Function to return the count of numbers` `// from 1 to n which are divisible by` `// all the numbers from 2 to 10` `function` `countNumbers(` `$n` `)` `{` ` ` `return` `(int)(` `$n` `/ 2520);` `}` `// Driver code` `$n` `= 3000;` `echo` `(countNumbers(` `$n` `));` `// This code is contributed` `// by Code_Mech.` `?>` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the count of numbers` `// from 1 to n which are divisible by` `// all the numbers from 2 to 10` `function` `countNumbers(n)` `{` ` ` `return` `(n / 2520);` `}` `// Driver code` `var` `n = 3000;` ` ` `// Function call` `document.write(Math.round(countNumbers(n)));` `// This code is contributed by Ankita saini` `</script>` |

**Output:**

1

**Time Complexity**: O(1)**Auxiliary Space:** O(1)