# Smallest odd number with N digits

• Last Updated : 22 Jun, 2022

Given a number N. The task is to find the smallest N digit ODD number.
Examples:

Input: N = 1
Output: 1

Input: N = 3
Output: 101

Approach: There can be two cases depending on the value of N.
Case 1 : If N = 1 then answer will be 1.
Case 2 : If N != 1 then answer will be (10^(n-1)) + 1 because the series of the smallest odd numbers will go on like: 1, 11, 101, 1001, 10001, 100001, ….
Below is the implementation of the above approach:

## C++

 // C++ implementation of the above approach   #include using namespace std;   // Function to return smallest odd // with n digits int smallestOdd(int n) {     if (n == 1)         return 1;       return pow(10, n - 1) + 1; }   // Driver Code int main() {     int n = 4;     cout << smallestOdd(n);       return 0; }

## Java

 // Java implementation of the approach class Solution {       // Function to return smallest odd with n digits     static int smallestOdd(int n)     {         if (n == 1)             return 0;         return Math.pow(10, n - 1) + 1;     }       // Driver code     public static void main(String args[])     {         int n = 4;           System.out.println(smallestOdd(n));     } }

## Python3

 # Python3 implementation of the approach   # Function to return smallest even # number with n digits def smallestOdd(n) :       if (n == 1):         return 1     return pow(10, n - 1) + 1   # Driver Code n = 4 print(smallestOdd(n))   # This code is contributed by ihritik.

## C#

 // C# implementation of the approach using System; class Solution {       // Function to return smallest odd with n digits     static int smallestOdd(int n)     {         if (n == 1)             return 0;           return Math.pow(10, n - 1) + 1;     }       // Driver code     public static void Main()     {         int n = 4;           Console.Write(smallestOdd(n));     } }



## Javascript



Output:

1001

Time Complexity: O(log n).
Auxiliary Space: O(1)

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