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fix(curriculum): clarify leap year in user story (#57700)

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Krzysztof G. 2024-12-23 09:19:25 +01:00 committed by GitHub
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curriculum/challenges/english/25-front-end-development/lab-leap-year-calculator/66c06fad3475cd92421b9ac2.md
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@ -15,12 +15,12 @@ Fulfill the user stories below and get all the tests to pass to complete the lab
1. Define a function called `isLeapYear` that takes a number as an argument.
2. Outside the function, declare a variable `year` that stores the value of the year you want to check.
3. Inside the function, use an `if/ else` statement or a ternary operator to check if the year is a leap year.
4. To check if the year is a leap year, fulfill the following conditions:
3. Inside the function, use an `if/ else` statement or a ternary operator to check if the year is a leap year.
4. To check if the year is a leap year, fulfill the following conditions:
- If the year is divisible by `4`, then it is a leap year.
- If the year is divisible by `100`, then it is not a leap year.
- If the year is divisible by `400`, then it is a leap year.
- Unless the year is also divisible by `100`, then it is not a leap year.
- Unless the year is also divisible by `400`, then it is a leap year.
5. If the year is a leap year, return `[year] is a leap year.`. Otherwise, return `[year] is not a leap year.`. You will replace `[year]` with the parameter defined in the `isLeapYear` function.
6. You should call the `isLeapYear` function with `year` as the argument and assign the result to a variable named `result`.