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Count numbers which are divisible by all the numbers from 2 to 10
  • Difficulty Level : Easy
  • Last Updated : 14 Apr, 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 <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

 

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