计算可以被2到10的所有数字整除的数字

给定整数N ，任务是找到1到N的数量计数，该数量可以被2到10的所有数字整除。

例子：

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 
 

方法：让我们将2到10的数字分解。

2 = 2 
3 = 3 
4 = 2² 
5 = 5 
6 = 2 * 3 
7 = 7 
8 = 2³ 
9 = 3² 
10 = 2 * 5 
 

如果一个数字可以被2到10的所有数字整除，则其因式分解应至少包含2的幂， 3的幂， 3的至少2的幂， 5的整数和7的至少1的幂。因此可以写成：

x * 2³ * 3² * 5 * 7 i.e. x * 2520. 
 

因此，任何可以被2520整除的数字都可以被2到10的所有数字整除。因此，此类数字的数量为N / 2520 。

下面是上述方法的实现：

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

Javascript

 
// 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
 

输出：

1