Smallest N digit number divisible by N

Given a positive integers N, the task is to find the smallest N digit number divisible by N.

Examples:

Input: N = 2
Output: 10
Explanation:
10 is the smallest 2-digit number which is divisible by 2.

Input: N = 3
Output: 102
Explanation:
102 is the smallest 3-digit number which is divisible by 3.

Naive Approach: The naive approach is to iterate from smallest N-digit number(say S) to largest N-digit number(say L). The first number between [S, L] divisible by N is the required result.



Below is the implementation of above approach:

C++

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// C++ program for the above approach
#include <iostream>
#include <math.h>
using namespace std;
  
// Function to find the smallest
// N-digit number divisible by N
void smallestNumber(int N)
{
    // Find largest n digit number
    int L = pow(10, N) - 1;
  
    // Find smallest n digit number
    int S = pow(10, N - 1);
  
    for (int i = S; i <= L; i++) {
  
        // If i is divisible by N,
        // then print i and return ;
        if (i % N == 0) {
  
            cout << i;
            return;
        }
    }
}
  
// Driver Code
int main()
{
    // Given Number
    int N = 2;
  
    // Function Call
    smallestNumber(N);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the smallest
// N-digit number divisible by N
static void smallestNumber(int N)
{
  
    // Find largest n digit number
    int L = (int) (Math.pow(10, N) - 1);
  
    // Find smallest n digit number
    int S = (int) Math.pow(10, N - 1);
  
    for (int i = S; i <= L; i++) 
    {
  
        // If i is divisible by N,
        // then print i and return ;
        if (i % N == 0
        {
            System.out.print(i);
            return;
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    // Given Number
    int N = 2;
  
    // Function Call
    smallestNumber(N);
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program for the above approach
  
# Function to find the smallest
# N-digit number divisible by N
def smallestNumber(N):
  
    # Find largest n digit number
    L = pow(10, N) - 1;
  
    # Find smallest n digit number
    S = pow(10, N - 1);
  
    for i in range(S, L): 
  
        # If i is divisible by N,
        # then print i and return ;
        if (i % N == 0): 
            print(i);
            return;
          
# Driver Code 
if __name__ == "__main__" :
      
    # Given number
    N = 2;
  
    # Function call
    smallestNumber(N)
  
# This code is contributed by rock_cool

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C#

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// C# program for the above approach
using System;
class GFG{
  
// Function to find the smallest
// N-digit number divisible by N
static void smallestNumber(int N)
{
  
    // Find largest n digit number
    int L = (int)(Math.Pow(10, N) - 1);
  
    // Find smallest n digit number
    int S = (int)Math.Pow(10, N - 1);
  
    for(int i = S; i <= L; i++) 
    {
         
       // If i is divisible by N,
       // then print i and return ;
       if (i % N == 0) 
       {
           Console.Write(i);
           return;
       }
    }
}
  
// Driver Code
public static void Main()
{
      
    // Given number
    int N = 2;
  
    // Function call
    smallestNumber(N);
}
}
  
// This code is contributed by Nidhi_biet

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Output:

10

Time Complexity: O(L – S), where L and S is the largest and smallest N-digit number respectively.

Efficient Approach: If the number divisible by N, then the number will be of the form N * X for some positive integer X.
Since it has to be smallest N-digit number, then X will be given by: \lceil \frac{10^{N-1}}{N} \rceil.
Therefore, the smallest number N-digit number is given by:  N*\lceil \frac{10^{N-1}}{N} \rceil

For Example:

For N = 3, the smallest 3-digit number is given by:
=>  3*\lceil \frac{10^{3-1}}{3} \rceil
=>  3*\lceil \frac{100}{3} \rceil
=>  3*\lceil \33.3 \rceil
=>  102

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <iostream>
#include <math.h>
using namespace std;
  
// Function to find the smallest
// N-digit number divisible by N
int smallestNumber(int N)
{
  
    // Return the smallest N-digit
    // number calculated using above
    // formula
    return N * ceil(pow(10, (N - 1)) / N);
}
  
// Driver Code
int main()
{
    // Given N
    int N = 2;
  
    // Function Call
    cout << smallestNumber(N);
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the smallest
// N-digit number divisible by N
static int smallestNumber(int N)
{
  
    // Return the smallest N-digit
    // number calculated using above
    // formula
    return (int) (N * Math.ceil(Math.pow(10, (N - 1)) / N));
}
  
// Driver Code
public static void main(String[] args)
{
    // Given N
    int N = 2;
  
    // Function Call
    System.out.print(smallestNumber(N));
}
}
  
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above approach
import math
  
# Function to find the smallest
# N-digit number divisible by N
def smallestNumber(N):
  
    # Return the smallest N-digit
    # number calculated using above
    # formula
    return N * math.ceil(pow(10, (N - 1)) // N);
  
# Driver Code
  
# Given N
N = 2;
  
# Function Call
print(smallestNumber(N));
  
# This code is contributed by Code_Mech

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C#

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// C# program for the above approach
using System;
class GFG{
  
// Function to find the smallest
// N-digit number divisible by N
static int smallestNumber(int N)
{
  
    // Return the smallest N-digit
    // number calculated using above
    // formula
    return (int) (N * Math.Ceiling(Math.Pow(10, (N - 1)) / N));
}
  
// Driver Code
public static void Main()
{
    // Given N
    int N = 2;
  
    // Function Call
    Console.Write(smallestNumber(N));
}
}
  
// This code is contributed by Code_Mech

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Output:

10

Time Complexity: O(1)

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