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 
 

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++ 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 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 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# 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 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.
?>

<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>

1

Time Complexity: O(1), since there is a basic arithmetic operation that takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.