Rule

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

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

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)

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;
}

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

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") 

oh of course there is

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

(do take a look at the download stats)

just check the least significant bit smh my head

https://codegolf.stackexchange.com/q/275739/88192

I thought they were going to turn into Saddam Husseins.

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