A common year in the modern Gregorian Calendar consists of 365 days. In reality, Earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: 1) The year must be divisible by 4 2) If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400 Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines the number of days in February for that year

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A common year in the modern Gregorian Calendar consists of 365 days. In reality, Earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are:

1) The year must be divisible by 4

2) If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400

Some example leap years are 1600, 1712, and 2016.

Write a program that takes in a year and determines the number of days in February for that year.

My Code:

def days_in_feb(user_year):
    is_leap_year = 29
    not_leap_year = 28
    if user_year % 4 ==0:
        return is_leap_year
    elif user_year % 4 ==0 and user_year % 100 == 0 and user_year % 400 == 0:
        return is_leap_year
    else:  
        return(not_leap_year)
    
if __name__ == '__main__':
    # Type your code here. Your code must call the function.
    user_year = int(input())
    days = days_in_feb(user_year)
    print(user_year,'has', days, 'days in February.')

Error: 

I get 7/10 points but I fail the 1900 question.

days_in_feb(1900)
Your output: days_in_feb(1900) incorrectly returned 29
The reason that I am confused is that if 1600, 1700, and 1800 return 29 then why would 1900 be different.

 

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i see the correction with my code.  However the error is now occuring on input 1600.

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