Difficulty: Medium
Correct Answer: Friday
Explanation:
Introduction / Context:
This problem tests your ability to track how weekdays shift from year to year, especially across leap years. In calendar aptitude, it is crucial to know how many days a year contributes modulo 7 to the day-of-week change from one year to the next.
Given Data / Assumptions:
Concept / Approach:
Each non-leap year has 365 days, which is 52 weeks + 1 day, so it shifts the weekday by +1. A leap year has 366 days, which is 52 weeks + 2 days, shifting the weekday by +2. We count how many leap and non-leap years occur between the given years and add their contributions modulo 7 to determine the weekday shift from 2006 to 2010.
Step-by-Step Solution:
Step 1: List the years from 2006 up to (but not including) 2010: 2006, 2007, 2008, 2009.Step 2: Identify leap years among them. 2008 is divisible by 4 and is not a century year, so it is a leap year. The others (2006, 2007, 2009) are non-leap years.Step 3: Each non-leap year contributes a +1 day shift. Each leap year contributes a +2 day shift.Step 4: Total shift = (3 non-leap years * 1 day) + (1 leap year * 2 days) = 3 + 2 = 5 days.Step 5: Starting day is Sunday on 1 January 2006. Moving forward 5 days: Sunday → Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5).Step 6: Therefore, 1 January 2010 falls on a Friday.
Verification / Alternative check:
A quick cross-check is to note that 4 years is about 1461 days (365 + 365 + 366 + 365). Since 1461 mod 7 = 5, the weekday must shift by 5 days, which is exactly what we computed. Starting from Sunday and moving 5 days forward again gives Friday.
Why Other Options Are Wrong:
Sunday would imply no net shift, Monday a shift of 1 day, Tuesday a shift of 2 days, and Saturday a shift of 6 days. None of these match the correct 5-day shift caused by 3 non-leap years and 1 leap year in the interval.
Common Pitfalls:
Final Answer:
The day of the week on 1 January 2010 was Friday.
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