Calendar arithmetic — If 14 October 2005 was a Friday, then what day of the week was 14 October 2006?

Introduction / Context:Day-of-week problems rely on year lengths and leap-year adjustments. We move exactly one calendar year forward from 14 Oct 2005 to 14 Oct 2006.

Given Data / Assumptions:

• 14 Oct 2005 is Friday.
• 2006 is not a leap year.
• From a date to the same date next year, the weekday advances by 1 day if the year in between has 365 days (non-leap), or by 2 days if 366 (leap).

Concept / Approach:Count the day shift over one year. 2006 (the year reached) does not matter; what matters is whether the elapsed year length between the dates is 365 or 366.

Step-by-Step Solution:Elapsed length from 14 Oct 2005 to 14 Oct 2006 = 365 days.A 365-day shift advances the weekday by +1.Friday + 1 = Saturday.

Verification / Alternative check:Since neither Feb 29 nor a leap-day crossing occurs in this one-year span (2006 is not leap), the +1 shift is consistent.

Why Other Options Are Wrong:Tuesday/Thursday/Friday correspond to +3/–1/0 shifts respectively; the correct shift here is +1.

Common Pitfalls:Looking at the property of the destination year (2006) rather than the elapsed year length; the rule depends on the 365/366-day span.

Final Answer:Saturday.