Exercise 2 solution

    import calendar
    from datetime import date

    born = date(2024, 2, 29)

    def birthday_in(born, year):
        return date(year, born.month, min(born.day, calendar.monthrange(year, born.month)[1]))

    for year in range(2025, 2030):
        b = birthday_in(born, year)
        print(f"{year}: {b:%a %d %b %Y}  leap={calendar.isleap(year)}")
    print("leap years 1896-1912:", [y for y in range(1896, 1913) if calendar.isleap(y)])
    print("count 1900-2100 inclusive:", calendar.leapdays(1900, 2101))

Output (run, and exactly as shown):

    2025: Fri 28 Feb 2025  leap=False
    2026: Sat 28 Feb 2026  leap=False
    2027: Sun 28 Feb 2027  leap=False
    2028: Tue 29 Feb 2028  leap=True
    2029: Wed 28 Feb 2029  leap=False
    leap years 1896-1912: [1896, 1904, 1908, 1912]
    count 1900-2100 inclusive: 49

Explanation: 29 February has to be mapped to a real date in non-leap years;
this solution uses 28 February (some legal systems use 1 March, so the policy
is a decision, not a fact of nature). 1900 is skipped because years divisible
by 100 are leap years only when also divisible by 400, which is why 1900 and
2100 are not leap years. calendar.leapdays(y1, y2) counts leap years from y1
up to but not including y2, so the end value is 2101.