Exercise 3 solution

    from datetime import datetime

    dt = datetime(2026, 3, 5, 9, 7, 4)

    line = dt.strftime("%a %d %b %Y, week %V, day %j")
    print(line)

    # Check: parse the date part back and compare
    date_part = line.split(",")[0]
    back = datetime.strptime(date_part, "%a %d %b %Y")
    print(back.date() == dt.date())

Output (run, and exactly as shown):

    Thu 05 Mar 2026, week 10, day 064
    True

Explanation: literal text in a format string is copied through unchanged, so
", week " and ", day " need no escaping; only the percent codes are
replaced. %V is the ISO week number and %j the day of the year. The check
splits at the first comma and parses the left part with the matching format.
Note that the weekday name in the text is read by strptime but does not change
which date it is: the day, month and year decide that. Names depend on the
locale, so this exact output assumes an English locale.