**Leap Year Checker in Python**

In this article, we will explore a Python code snippet that checks whether a year is a leap year or not. The code is quite concise and uses some clever logic to achieve this.

**The Code**

```
x = 2021
print((x % 4 == 0 and x % 100 != 0) or x % 400 == 0)
```

Let's break down what this code does:

### What is a Leap Year?

A leap year is exactly 366 days long, instead of the usual 365 days. Leap years are necessary to keep our calendar aligned with the Earth's orbit around the Sun.

### The Rules

To determine whether a year is a leap year, the following rules apply:

**Rule 1**: The year must be evenly divisible by 4.**Rule 2**: If the year is a century year (i.e., it is divisible by 100), then it must also be divisible by 400.

### The Code Explained

The code uses the modulo operator (`%`

) to check whether the year (`x`

) satisfies the rules.

`x % 4 == 0`

checks if the year is divisible by 4 (Rule 1).`x % 100 != 0`

checks if the year is not a century year (i.e., it is not divisible by 100).`x % 400 == 0`

checks if the year is divisible by 400 (Rule 2 for century years).

The code uses a logical `and`

operator to combine the first two conditions, and a logical `or`

operator to combine the result with the third condition.

### How it Works

When we run the code with `x = 2021`

, the output will be `False`

, because 2021 is not a leap year.

Here's a step-by-step breakdown:

`x % 4 == 0`

is`False`

, because 2021 is not divisible by 4.`x % 100 != 0`

is`True`

, because 2021 is not a century year.- The
`and`

operator short-circuits, and the expression evaluates to`False`

. - The
`or`

operator then evaluates the third condition,`x % 400 == 0`

, which is also`False`

. - The final output is
`False`

, indicating that 2021 is not a leap year.

### Try it Yourself!

Modify the value of `x`

to test different years, and see how the code works for yourself!