Build a Leap Year Calculator - Build a Leap Year Calculator

JavaScript
1

Tell us what’s happening:

I don’t understand what is wrong here, I checked code many times but it doesn’t helps. Please help me

Your code so far

function isLeapYear(year) {
  if((year % 4) === 0 && (year % 400) === 0) {
    return `${year} is a leap year.`
  } else if((year % 4) === 0 && (year%100) === 0) {
    return `${year} is not a leap year.`
  }
}

let year = 2024;
let result = isLeapYear(year);
console.log(result);

year = 2000;
result = isLeapYear(year);
console.log(result);

year = 1900;
result = isLeapYear(year);
console.log(result);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/131.0.0.0 Safari/537.36

Challenge Information:

Build a Leap Year Calculator - Build a Leap Year Calculator

2

You’re almost there but your logic doesn’t account for any case where the year is divisible by 4 but not divisible by either 100 or 400 (e.g. 2024, which is one of the cases which you’re failing, as your function returns undefined in this case).

3

you are also not accounting for years that are not divisible by 4. like 1997

1 Like
4

how can I think like you that how many cases I need to create to write a robust condition ? Is there any mathematical formula ?

5

Yes, doing it now. Also I want to ask if there is any mathematical formula to find how many cases I need to create to write robust code ?

6

You have the three cases described in the instructions:

  • If the year is divisible by 4, 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.

To which you would need to add the one that is implicit: when the year is not divisible by 4

7

done, test passed,

solution removed by moderator

thank you @igorgetmeabrain and @ILM

1 Like