so did someone draw this by hand or was it a filter

Using Haskell you can write it way more concise:

iseven :: Int -> Bool
iseven 0 = True
iseven 1 = False
iseven 2 = True
iseven 3 = False
iseven 4 = True
iseven 5 = False
iseven 6 = True
iseven 7 = False
iseven 8 = True
...

However, we can be way smarter by only defining the 2 base cases and then a recursive definition for all other numbers:

iseven :: Int -> Bool
iseven 0 = True
iseven 1 = False
iseven n = iseven (n-2)

It's having a hard time with negative numbers, but honestly that's quite a mood

a wise programmer knows to always ask the question "can i solve this problem in python using metaprogramming?" in this instance, the answer is yes:

def is_even(n: int):
    s = "def is_even_helper(number: int):\n"
    b = True
    for i in range(0, abs(n)+2):
        s += f"\tif (abs(number) == {i}): return {b}\n"
        b = not b
    exec(s)
    return locals().get("is_even_helper")(n)

oh of course there is

https://www.npmjs.com/package/is-even

(do take a look at the download stats)

I thought they were going to turn into Saddam Husseins.

=if((number/2)-round(number/2,0)=0,true,false)

return true

is correct around half of the time

import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None

Just divide the number into its prime factors and then check if one of them is 2.

Zero people in this post get the YanDev reference

Ask AI:

public static boolean isEven(int number) {
    // Handle negative numbers
    if (number < 0) {
        number = -number; // Convert to positive
    }
    
    // Subtract 2 until we reach 0 or 1
    while (number > 1) {
        number -= 2;
    }
    
    // If we reach 0, it's even; if we reach 1, it's odd
    return number == 0;
}

When you sacrifice memory for an O(1) algorithm.

In this case still O(n)

if (!(number & 1))

If number%2 == 0: return("Even")
Else: return("odd")