# Python – Find if given year is Leap Year in Georgian Calendar

## Python – Check if year is Leap Year in Georgian Calendar

In this tutorial, we will write a Python function to check if the given year is a leap year in Georgian Calendar.

Following are the criteria for an year to be Leap year.

• The year can be evenly divided by 4, is a leap year, unless:
• The year can be evenly divided by 100, it is NOT a leap year, unless:
• The year is also evenly divisible by 400. Then it is a leap year.

You may combine all these conditions into a single boolean expression.

Consider that we have our year in variable year.

To satisfy the first condition, “The year can be evenly divided by 4, is a leap year”, following condition has to be satisfied.

``year%4==0``

To satisfy the second condition, “The year can be evenly divided by 100, it is NOT a leap year,”, following condition has to be satisfied.

``year%100==0``

To satisfy the third condition, ” The year is also evenly divisible by 400″, following condition has to be satisfied.

``year%400==0``

By combining above three conditions, we get the following boolean expression.

``year%4==0 and not (year%100==0 and not year%400==0)``

unless in the given condition translates to and not and hence the and not operators in the above condition.

Following is the example program, in which we check if a given year is a leap year or not.

Python Program

``````def is_leap(year):
return year%4==0 and not (year%100==0 and not year%400==0)

year = int(input())

print(is_leap(year))``````
Copy

Output

``````2004
True``````

When you run the program, input() waits for your input. Enter a number and our program shall check if the given number is a leap year or not.

## Summary

In this tutorial, we learned how to check if given year is a leap year of Georgian Calendar or not.

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