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++ 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;
} |
// 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 |
# 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 |
// 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 |
<script> // Javascript program for the above approach // Function to find the smallest // N-digit number divisible by N function smallestNumber(N)
{ // Find largest n digit number
let L = Math.pow(10, N) - 1;
// Find smallest n digit number
let S = Math.pow(10, N - 1);
for (let i = S; i <= L; i++)
{
// If i is divisible by N,
// then print i and return ;
if (i % N == 0)
{
document.write(i);
return ;
}
}
} // Driver code // Given Number let N = 2; // Function Call smallestNumber(N); // This code is contributed by divyeshrabadiya07 </script> |
Output:
10
Time Complexity: O(L – S), where L and S is the largest and smallest N-digit number respectively.
Auxiliary Space: O(1)
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:
For Example:
For N = 3, the smallest 3-digit number is given by:
=>=>
=>
=> 102
Below is the implementation of the above approach:
// 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;
} |
// 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 |
# 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 |
// 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 |
<script> // Javascript program for the above approach
// Function to find the smallest
// N-digit number divisible by N
function smallestNumber(N)
{
// Return the smallest N-digit
// number calculated using above
// formula
return N * Math.ceil(Math.pow(10, (N - 1)) / N);
}
// Given N
let N = 2;
// Function Call
document.write(smallestNumber(N));
// This code is contributed by divyesh072019.
</script> |
Output:
10
Time Complexity: O(log(N))
Auxiliary Space: O(1)