# Smallest N digit number divisible by all possible prime digits

Given an integer N, the task is to find the smallest N digit number divisible by all possible prime digits, i.e, 2, 3, 5 and 7. Print -1 if no such number is possible.
Examples:

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

Approach:
Follow the steps given below to solve the problem:

• Since all the four numbers 2, 3, 5, 7 are prime it means N will also be divisible by their product 2 × 3 × 5 × 7 = 210
• For N < 3, no such number exists. So, print -1.
• For N = 3, the answer will be 210.
• For N > 3, the following computation needs to be done:
• Find Remainder R = 10N-1 % N.
• Add 210 – R to 10N-1.

Below is the implementation of the above approach:

## 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 `

Output:

```10080
```

Time complexity: O(logN)
Auxiliary Space: O(1)

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