Leap year

Question

Given the year number. You need to check if this year is a leap year. If it is, print LEAP, otherwise print COMMON.
The rules in Gregorian calendar are as follows:

a year is a leap year if its number is exactly divisible by 4 and is not exactly divisible by 100
a year is always a leap year if its number is exactly divisible by 400
Warning. The words LEAP and COMMON should be printed all caps.

thats my code please can someone explain why it doesn't work properly?
here is where you can check my code: https://snakify.org/en/lessons/if_then_else_conditions/problems/leap_year/

a = int(input())
if abs(a % 4 == 0) or a 4 ==0:
print("LEAP")
elif abs(a / 400):
print("COMMON")

Answer 1

#In 2nd line of code 'and' will be used not or and the code will be

a = int(input())
if abs(a % 4 == 0) and a%100 != 0:
print("LEAP")
elif abs(a / 400):
print("COMMON")
#Just copy and paste it in Sololearn Codes playground and see it's work

Answer 2

a%4==0 and not a%100==0 or a%400==0: leap
Else: Common

Answer 3

def is_leap_year(year):
"""Check whether a year is a leap year.

A leap year is divisible by 4 but not if
divisible by 100 unless also divisible by 400.
"""
return not year % 400 or not year % 4 and year % 100

Answer 4

year = int(input())
if abs(year % 4 == 0) and abs(year % 100 != 0):
print("Leap year")
elif abs(year // 400 ):
print("Not a leap year")

"""
cant find whats wrong but here

"'"

Answer 5

CancersGr.t
1) output should be LEAP / COMMON.
2) your categories are reversed
3) control flow is wrong. Either reverse order (check 400 first etc). Or combine conditions with "and" like in the description
4) also for 400 you have to use modulo

Additionally you don't need abs()