# 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 = 5Output:10080Explanation:10080 is the smallest five-digit number that is divisible by 2, 3, 5 and 7.Input:N = 3Output: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 = 10**.^{N-1}% N - Add
**210 – R**to**10**.^{N-1}

- Find Remainder

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h>` `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

`<script>` `// 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.` `</script>` |

**Output:**

10080

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