# 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 = 3000
Output: 1
2520 is the only number below 3000 which is divisible by all the numbers from 2 to 10.

Input: N = 2000
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

Output:

```1
```

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