Given an integer N, the task is to find an N-digit number such that it is not divisible by any of its digits.
Note: There can be multiple answers for each value of N.
Examples:
Input: N = 4
Output: 6789
Explanation:
As the number 6789 is not divisible by any of its digits, it is 6, 7, 8 and 9 and it is also a four-digit number. Hence, it can be the desired number.Input: N = 2
Output: 57
Explanation:
As the number 57 is not divisible by any of its digits, it is 5 and 7 and it is also a 2-digit number. Hence, it can be the desired number.
Approach: The key observation in the problem is that 2 and 3 are those numbers that don’t divide each other. Also, the numbers “23, 233, 2333, …” are not divisible by neither 2 nor 3. Hence, for any N-digit number, the most significant digit will be 2 and the rest of the digits will be 3 to get the desired number.
Algorithm:
- Check if the value of the N is equal to 1, then there is no such number is possible, hence return -1.
- Otherwise, initialize a variable num, to store the number by 2.
- Run a loop from 1 to N and then, for each iteration, multiply the number by 10 and add 3 to it.
num = (num * 10) + 3
Below is the implementation of the above approach:
// C++ implementation to find a // N-digit number such that the number // it is not divisible by its digits #include <bits/stdc++.h> using namespace std;
typedef long long int ll;
// Function to find the number // such that it is not divisible // by its digits void solve(ll n)
{ // Base Cases
if (n == 1)
{
cout << -1;
}
else {
// First Digit of the
// number will be 2
int num = 2;
// Next digits of the numbers
for (ll i = 0; i < n - 1; i++) {
num = (num * 10) + 3;
}
cout << num;
}
} // Driver Code int main()
{ ll n = 4;
// Function Call
solve(n);
} |
// Java implementation to find a // N-digit number such that the number // it is not divisible by its digits import java.io.*;
public class GFG {
long ll;
// Function to find the number
// such that it is not divisible
// by its digits
static void solve( long n)
{
// Base Cases
if (n == 1 )
{
System.out.println(- 1 );
}
else {
// First Digit of the
// number will be 2
int num = 2 ;
// Next digits of the numbers
for ( long i = 0 ; i < n - 1 ; i++) {
num = (num * 10 ) + 3 ;
}
System.out.println(num);
}
}
// Driver Code
public static void main (String[] args)
{
long n = 4 ;
// Function Call
solve(n);
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation to find a # N-digit number such that the number # it is not divisible by its digits # Function to find the number # such that it is not divisible # by its digits def solve(n) :
# Base Cases
if (n = = 1 ) :
print ( - 1 );
else :
# First Digit of the
# number will be 2
num = 2 ;
# Next digits of the numbers
for i in range (n - 1 ) :
num = (num * 10 ) + 3 ;
print (num);
# Driver Code if __name__ = = "__main__" :
n = 4 ;
# Function Call
solve(n);
# This code is contributed by AnkitRai01 |
// C# implementation to find a // N-digit number such that the number // it is not divisible by its digits using System;
class GFG {
long ll;
// Function to find the number
// such that it is not divisible
// by its digits
static void solve( long n)
{
// Base Cases
if (n == 1)
{
Console.WriteLine(-1);
}
else {
// First Digit of the
// number will be 2
int num = 2;
// Next digits of the numbers
for ( long i = 0; i < n - 1; i++) {
num = (num * 10) + 3;
}
Console.WriteLine(num);
}
}
// Driver Code
public static void Main(String[] args)
{
long n = 4;
// Function Call
solve(n);
}
} // This code is contributed by sapnasingh4991 |
<script> //Javascript implementation to find a // N-digit number such that the number // it is not divisible by its digits // Function to find the number // such that it is not divisible // by its digits function solve(n)
{ // Base Cases
if (n == 1)
{
document.write( -1);
}
else {
// First Digit of the
// number will be 2
var num = 2;
// Next digits of the numbers
for ( var i = 0; i < n - 1; i++) {
num = (num * 10) + 3;
}
document.write( num);
}
} // Given N var n = 4;
// Function Call solve(n); // This code is contributed by SoumikMondal </script> |
2333
Performance Analysis:
- Time Complexity: O(N).
- Auxiliary Space: O(1).