## 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.

- The year can be evenly divided by 100, it is NOT a leap year, unless:

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`

Run 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`

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

`year%400==0`

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

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

Run 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))
```

**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 of Python Examples, we learned how to check if given year is a leap year of Georgian Calendar or not.