Beautiful Non Divisible Number

Given an integer N find an N digit positive number X such that GCD of X and any of its digit is not greater than 1. Print -1 if the answer does not exist.

Examples:

Input: N = 5 
Output: 57777
Explanation: 57777 has in total 5 digits in which gcd(57777, 5) = 1 and gcd(57777, 7) = 1

Input: N = 2
Output: 57
Explanation: 57 has in total 2 digits in which gcd(57, 5)=1 and gcd(57, 7) = 1 

Approach: To solve the problem, follow the below idea:

So, the main idea here is that we need to take in total N digits in a number X which should be > 0 and at the same time their gcd with X should be 1. If we try to take every time first digit as 5 in X and the remaining digit i.e N – 1 as 7 we are done with our target. As, this number has each digit > 0 and there gcd with s is always bound to be 1.

Step-by-step algorithm:

• Check if N == 1, then it is impossible to construct the answer.
• Initialize a string X of length N having all characters as ‘7’.
• Change the first character to 5.
• Print the string X as the final answer.

Below is the implementation of the algorithm:

C++

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N = 13;

    // If N == 1, then it is impossible to construct the answer
    if (N == 1) { 
        cout << "-1\n";
    }

    // Construct a number having N digits, all equal to 7
    string X(N, '7');

    // Change the first digit to 5
    X[0] = '5';
    cout << X << "\n";

    return 0;
}

Java

public class Main {
    public static void main(String[] args) {
        int N = 13;

        // If N == 1, then it is impossible to construct the answer
        if (N == 1) {
            System.out.println("-1");
        }

        // Construct a number having N digits, all equal to 7
        StringBuilder X = new StringBuilder("7".repeat(N));

        // Change the first digit to 5
        X.setCharAt(0, '5');
        System.out.println(X);
    }
}

// This code is contributed by rambabuguphka

Python

def GFG(N):
    # If N == 1, then it is impossible to construct the answer
    if N == 1:
        return "-1"
    # Construct a number having N digits all equal to 7
    X = "7" * N
    # Change the first digit to 5
    X = "5" + X[1:]
    return X

# Main function
def main():
    N = 13
    X = GFG(N)
    print(X)

main()

C#

using System;

class Program
{
    static void Main()
    {
        int N = 13;

        // If N == 1, then it is impossible to construct the answer
        if (N == 1)
        {
            Console.WriteLine("-1");
        }
        else
        {
            // Construct a string having N digits, all equal to '7'
            string X = new String('7', N);

            // Change the first digit to '5'
            X = '5' + X.Substring(1);

            Console.WriteLine(X);
        }
    }
}

Javascript

function GFG(N) {
    // If N == 1, then it is impossible to the construct the answer
    if (N === 1) {
        return "-1";
    }
    // Construct a number having N digits
    // all equal to 7
    let X = "7".repeat(N);
    // Change the first digit to 5
    X = "5" + X.substring(1);
    return X;
}
// Main function
function main() {
    const N = 13;
    const X = GFG(N);
    console.log(X);
}
main();

Output

5777777777777

Time Complexity: O(N), where N is the number of digits of number X.
Auxiliary Space: O(1)