Difficulty: Easy
Correct Answer: Sunday
Explanation:
Introduction / Context:Determining the weekday for the last day of a year is a classic calendar arithmetic problem. We use the facts that a common (non-leap) year has 365 days (which is 52 weeks + 1 day), and a leap year has 366 days (52 weeks + 2 days). This shifts the weekday of the last day relative to the first day of the year.
Given Data / Assumptions:
Concept / Approach:In any common year (365 days), the weekday advances by +1 from January 1 to December 31. If we know January 1, 2006, then December 31, 2006 is the same weekday as January 1, 2006 (because there are 365 = 7 * 52 + 1 days, and the +1 shift from Jan 1 to Jan 2 accumulates so that Dec 31 matches Jan 1). Alternatively, recall (or compute) that January 1, 2006 was Sunday.
Step-by-Step Solution:
Check leap: 2006 % 4 = 2 ⇒ not a leap year.For a common year: Dec 31 has the same weekday as Jan 1.Known reference (or computed via doomsday/offset): Jan 1, 2006 = Sunday.Therefore, Dec 31, 2006 = Sunday.Verification / Alternative check:
Use any standard perpetual calendar or doomsday algorithm: 2006 anchor confirms Sunday for 2006-01-01 and 2006-12-31.Why Other Options Are Wrong:
Monday/Wednesday/Thursday contradict the one-day shift pattern and the known 2006 reference.Common Pitfalls:
Confusing leap vs common year (a leap year would shift the last day two weekdays ahead of Jan 1).Final Answer:Sunday
Discussion & Comments