Exercise 1 solution

    import calendar
    from datetime import date

    def add_months(d, n):
        year, month = divmod(d.year * 12 + d.month - 1 + n, 12)
        month += 1
        day = min(d.day, calendar.monthrange(year, month)[1])
        return date(year, month, day)

    start = date(2026, 1, 31)
    step = start
    print("n  direct      chained")
    for n in range(0, 5):
        direct = add_months(start, n)
        if n:
            step = add_months(step, 1)
        flag = "" if direct == step else "  <-- differs"
        print(f"{n}  {direct}  {step}{flag}")

Output (run, and exactly as shown):

    n  direct      chained
    0  2026-01-31  2026-01-31
    1  2026-02-28  2026-02-28
    2  2026-03-31  2026-03-28  <-- differs
    3  2026-04-30  2026-04-28  <-- differs
    4  2026-05-31  2026-05-28  <-- differs

Explanation: clamping loses information. Once 31 January has become 28
February, the original day number is gone, so adding one more month gives the
28th. Adding the whole number of months to the original date in one step keeps
the 31 and clamps only at the end. Rule: always add from the original date,
never from an already-clamped result.