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

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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++

// 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:

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.

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