能被所有可能的素数整除的最小 N 位数

给定一个整数N ，任务是找到可以被所有可能的素数整除的最小N位数，即2, 3, 5和7 。如果没有这样的数字，则打印-1 。
例子：

Input: N = 5 
Output: 10080 
Explanation: 10080 is the smallest five-digit number that is divisible by 2, 3, 5 and 7.
Input: N = 3 
Output: 210

方法：
请按照以下步骤解决问题：

• 由于所有四个数字2, 3, 5, 7都是素数，这意味着N也可以被它们的乘积整除2 × 3 × 5 × 7 = 210
• 对于N < 3 ，不存在这样的数字。所以，打印-1 。
• 对于N = 3 ，答案将是210 。
• 对于N > 3 ，需要进行以下计算：
  • 求余数R = 10 ^N-1 % N。
  • 将210 – R添加到10 ^N-1 。

下面是上述方法的实现：

C++

// C++ implementation of the above approach 
#include 
using namespace std;
 
// Function to find the minimum number of 
// n digits divisible by all prime digits
void minNum(int n)
{
    if (n < 3)
        cout << -1;
    else
        cout << (210 * ((int)(pow(10, n - 1) /
                                   210) + 1));
}
 
// Driver Code
int main()
{
    int n = 5;
    minNum(n);
    return 0;
}
 
// This code is contributed by amal kumar choubey

Java

// Java implementation of the above approach
class GFG{
     
// Function to find the minimum number of
// n digits divisible by all prime digits
static void minNum(int n)
{
    if(n < 3)
        System.out.println(-1);
    else
        System.out.println(210 * (
            (int)(Math.pow(10, n - 1) / 210) + 1));
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5;
    minNum(n);
}
}
 
// This code is contributed by Stuti Pathak

Python3

# Python3 implementation of the above approach
from math import *
  
# Function to find the minimum number of
# n digits divisible by all prime digits.
def minNum(n):
    if n < 3:
        print(-1)
    else:
        print(210 * (10**(n-1) // 210 + 1))
  
# Driver Code
n = 5
minNum(n)

C#

// C# implementation of the above approach
using System;
 
class GFG{
 
// Function to find the minimum number of
// n digits divisible by all prime digits
static void minNum(int n)
{
    if (n < 3)
        Console.WriteLine(-1);
    else
        Console.WriteLine(210 *
           ((int)(Math.Pow(10, n - 1) / 210) + 1));
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 5;
    minNum(n);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

// Javascript implementation of the above approach 
 
// Function to find the minimum number of 
// n digits divisible by all prime digits
function minNum(n)
{
    if (n < 3)
        document.write( -1);
    else
        document.write((210 * (parseInt(Math.pow(10, n - 1) /
                                   210) + 1)));
}
 
// Driver Code
var n = 5;
minNum(n);
 
// This code is contributed by rrrtnx.

输出：

10080

时间复杂度： O(logN)
辅助空间： O(1)