Given a positive integer **N**, the task is to check if the given number **N** has **at least 1** odd divisor from the range **[2, N – 1]** or not. If found to be true, then print **“Yes”**. Otherwise, print **“No”**.

**Examples:**

Input:N = 10Output:YesExplanation:

10 has 5 as the odd divisor. Therefore, print Yes.

Input:N = 8Output:No

**Approach:** The idea to solve the given problem is to iterate through all possible odd divisors over the range **[3, sqrt(N)]** and if there exists any such divisor, then print **“Yes”**. Otherwise, print **“No”**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check whether N` `// has at least one odd divisor` `// not exceeding N - 1 or not` `string oddDivisor(` `int` `N)` `{` ` ` `// Stores the value of N` ` ` `int` `X = N;` ` ` `// Reduce the given number` ` ` `// N by dividing it by 2` ` ` `while` `(N % 2 == 0) {` ` ` `N /= 2;` ` ` `}` ` ` `for` `(` `int` `i = 3; i * i <= X; i += 2) {` ` ` `// If N is divisible by` ` ` `// an odd divisor i` ` ` `if` `(N % i == 0) {` ` ` `return` `"Yes"` `;` ` ` `}` ` ` `}` ` ` `// Check if N is an odd divisor after` ` ` `// reducing N by dividing it by 2` ` ` `if` `(N != X)` ` ` `return` `"Yes"` `;` ` ` `// Otherwise` ` ` `return` `"No"` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 10;` ` ` `// Function Call` ` ` `cout << oddDivisor(N);` ` ` `return` `0;` `}` |

## Java

`/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to check whether N` ` ` `// has at least one odd divisor` ` ` `// not exceeding N - 1 or not` ` ` `public` `static` `String oddDivisor(` `int` `N)` ` ` `{` ` ` ` ` `// Stores the value of N` ` ` `int` `X = N;` ` ` `// Reduce the given number` ` ` `// N by dividing it by 2` ` ` `while` `(N % ` `2` `== ` `0` `) {` ` ` `N /= ` `2` `;` ` ` `}` ` ` `for` `(` `int` `i = ` `3` `; i * i <= X; i += ` `2` `) {` ` ` `// If N is divisible by` ` ` `// an odd divisor i` ` ` `if` `(N % i == ` `0` `) {` ` ` `return` `"Yes"` `;` ` ` `}` ` ` `}` ` ` `// Check if N is an odd divisor after` ` ` `// reducing N by dividing it by 2` ` ` `if` `(N != X) {` ` ` `return` `"Yes"` `;` ` ` `}` ` ` ` ` `// Otherwise` ` ` `return` `"No"` `;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `10` `;` ` ` ` ` `// Function Call` ` ` `System.out.println(oddDivisor(N));` ` ` `}` `}` `// This code is contributed by aditya7409.` |

## Python3

`# Python program for the above approach` `# Function to check whether N` `# has at least one odd divisor` `# not exceeding N - 1 or not` `def` `oddDivisor(N):` ` ` ` ` `# Stores the value of N` ` ` `X ` `=` `N` ` ` ` ` `# Reduce the given number` ` ` `# N by dividing it by 2` ` ` `while` `(N ` `%` `2` `=` `=` `0` `):` ` ` `N ` `/` `/` `=` `2` ` ` ` ` `i ` `=` `3` ` ` `while` `(i ` `*` `i <` `=` `X):` ` ` ` ` `# If N is divisible by` ` ` `# an odd divisor i` ` ` `if` `(N ` `%` `i ` `=` `=` `0` `):` ` ` `return` `"Yes"` ` ` `i ` `+` `=` `2` ` ` ` ` `# Check if N is an odd divisor after` ` ` `# reducing N by dividing it by 2` ` ` `if` `(N !` `=` `X):` ` ` `return` `"Yes"` ` ` ` ` `# Otherwise` ` ` `return` `"No"` ` ` `# Driver Code` `N ` `=` `10` `# Function Call` `print` `(oddDivisor(N))` `# This code is contributed by shubhamsingh10` |

## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function to check whether N` ` ` `// has at least one odd divisor` ` ` `// not exceeding N - 1 or not` ` ` `public` `static` `string` `oddDivisor(` `int` `N)` ` ` `{` ` ` `// Stores the value of N` ` ` `int` `X = N;` ` ` `// Reduce the given number` ` ` `// N by dividing it by 2` ` ` `while` `(N % 2 == 0) {` ` ` `N /= 2;` ` ` `}` ` ` `for` `(` `int` `i = 3; i * i <= X; i += 2) {` ` ` `// If N is divisible by` ` ` `// an odd divisor i` ` ` `if` `(N % i == 0) {` ` ` `return` `"Yes"` `;` ` ` `}` ` ` `}` ` ` `// Check if N is an odd divisor after` ` ` `// reducing N by dividing it by 2` ` ` `if` `(N != X) {` ` ` `return` `"Yes"` `;` ` ` `}` ` ` `// Otherwise` ` ` `return` `"No"` `;` ` ` `}` ` ` `// Driver Code` ` ` `static` `public` `void` `Main()` ` ` `{` ` ` `int` `N = 10;` ` ` `// Function Call` ` ` `Console.Write(oddDivisor(N));` ` ` `}` `}` `// This code is contributed by sanjoy_62.` |

**Output:**

Yes

**Time Complexity:** O(√N)**Auxiliary Space:** O(1)

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