Calendar arithmetic — If 14 October 2005 was a Friday, then what day of the week was 14 October 2009?

Introduction / Context:We move four years ahead from the same calendar date, accounting for leap years to determine the net weekday shift.

Given Data / Assumptions:

• Start: 14 Oct 2005 (Friday).
• Years traversed: 2006, 2007, 2008 (leap), 2009.
• Non-leap year shift = +1 day; leap year shift = +2 days (because 366 ≡ 1 mod 7 + an extra day).

Concept / Approach:Add up all yearly shifts between the two dates.

Step-by-Step Solution:2005→2006: +12006→2007: +12007→2008: +1 (to 2008-10-14, the leap effect is already captured in full years)2008→2009: +2? No—careful: the +2 applies when spanning a leap year entirely. Across 2008→2009, the elapsed year (2008–2009) has 366 days, so shift +2.Total shift = 1 + 1 + 2 + 1 = +5.Friday + 5 → Wednesday.

Verification / Alternative check:+5 means move 2 days backward from Friday to get Wednesday (since +7 ≡ 0). Counting forward: Fri→Sat(1)→Sun(2)→Mon(3)→Tue(4)→Wed(5).

Why Other Options Are Wrong:Tuesday/Thursday/Friday correspond to +4/+6/+0 shifts respectively; our net is +5.

Common Pitfalls:Ignoring the leap-year extra day or miscounting the number of leap years between dates.

Final Answer:Wednesday.