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

• Difficulty Level : Easy
• Last Updated : 12 Jul, 2021

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

Input: N = 2000
Output:

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

2 = 2
3 = 3
4 = 22
5 = 5
6 = 2 * 3
7 = 7
8 = 23
9 = 32
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 * 23 * 32 * 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 ``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

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## Javascript

 ``
Output:
`1`

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

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