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.
or divide the number by two and if the remainder is greater than
-(4^34)
but less than
70 - (((23*3*4)/2)/2)
then
true
What if the remainder is greater than the first, but not less than the latter?
Like, for example, 1?
Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I’ve only checked up to 4194304 to make sure this works, so if you need bigger numbers, you’ll have to validate on your own.
I remember coding actionscript in Flash and using modulo (%) to determine if a number was even or odd. It returns the remainder of the number divided by 2 and if it equals anything other than 0 then the number is odd.
Zero people in this post get the YanDev reference
so did someone draw this by hand or was it a filter
