Smallest odd number with N digits Last Updated : 22 Jun, 2022 Comments Improve Suggest changes Like Article Like Report 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 <bits/stdc++.h> 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)); } } PHP <?php // PHP implementation of the approach // Function to return smallest even // number with n digits function smallestOdd($n) { if ($n == 1) return 1; return pow(10, $n - 1) + 1; } // Driver Code $n = 4; echo smallestOdd($n); // This code is contributed by ihritik ?> JavaScript <script> // Javascript implementation of the above approach // Function to return smallest odd // with n digits function smallestOdd(n) { if (n == 1) return 1; return Math.pow(10, n - 1) + 1; } // Driver Code var n = 4; document.write(smallestOdd(n)); // This code is contributed by rrrtnx. </script> Output: 1001 Time Complexity: O(log n). 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