X=2021 Print((x 4==0 And X 100 =0) Or X 400==0)

4 min read Jun 26, 2024
X=2021 Print((x 4==0 And X 100 =0) Or X 400==0)

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:

  1. Rule 1: The year must be evenly divisible by 4.
  2. 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:

  1. x % 4 == 0 is False, because 2021 is not divisible by 4.
  2. x % 100 != 0 is True, because 2021 is not a century year.
  3. The and operator short-circuits, and the expression evaluates to False.
  4. The or operator then evaluates the third condition, x % 400 == 0, which is also False.
  5. 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!