Let us go through all the methods using which we can solve this problem of finding odd and even numbers

## Method 1: Using the number – 2 method

def check_even_odd(n): if(n==0): return True elif(n==1): return False else: return check_even_odd(n-2) num = int(input("Enter the number to check whether the number is even or odd:")) if(check_even_odd(num)): print(num," is an even number") else: print(num," is an odd number")

**Output**

```
Enter the number to check whether the number is even or odd:256
256 is an even number
```

**Explanation:**

Here in this method, we simply first took the input from the user and then passed on that input to a function that checks for even or odd. Inside the function, we used conditionals. We checked if the number 0 then it is even and if it is 1 then it is odd. If none of the above then we subtract 2 from the given number, if after subtracting the output comes 0 or 1 then it automatically satisfies the given conditionals, if not then we again subtract. In this manner, we used recursion (a function called by that function).

## Method 2: Using the remainder method

def check_even_odd(n): if(n % 2 == 0): return True elif(n %2 != 0): return False else: return check_even_odd(n) num = int(input("Enter the number to check whether the number is even or odd:")) if(check_even_odd(num)): print(num," is an even number (using remainder method)") else: print(num," is an odd number (using remainder method)")

**Explanation**

Here we used the modulo operator to find the remainder and then used the conditional.

## Method 3: Using the binary (&) method

def check_even_odd(n): if n & 1: return False else: return True num = int(input("Enter the number to check whether the number is even or odd:")) if(check_even_odd(num)): print(num," is an even number (using binary AND method)") else: print(num," is an odd number (using binary AND method)")

**Explanation**

In this last method, we used the concept of binary AND. We passed the value in the function and performed the **AND **operation on n and 1, if that turns out to be true then we will return **False **and then will simply print that the number is odd. But if the number is even, then it will return **True**.

To conclude, in this article, we looked at three methods to find even or odd numbers using precision. We encourage you to share other different methods in the comments.