Calculating the Day of the Week from 2005 to 2009

This problem requires us to find the day of the week for a specific date (December 31, 2009) given the day of the week for an earlier date (December 31, 2005). We can solve this by calculating the number of 'odd days' between the two dates.

Understanding Odd Days and Leap Years

An 'odd day' is the number of days remaining after dividing the total number of days by 7 (since there are 7 days in a week). Each normal year has 365 days. \(365 \div 7 = 52\) weeks and \(1\) day remaining. So, a normal year has 1 odd day.

A leap year has 366 days. \(366 \div 7 = 52\) weeks and \(2\) days remaining. So, a leap year has 2 odd days. Leap years occur every four years, with the exception of years divisible by 100 but not by 400. We need to identify the leap years in the period we are considering.

Identifying Years and Odd Days between 31 Dec 2005 and 31 Dec 2009

We are moving from 31 December 2005 to 31 December 2009. This period covers the full years 2006, 2007, 2008, and 2009.

• From 31 Dec 2005 to 31 Dec 2006 is the year 2006.
• From 31 Dec 2006 to 31 Dec 2007 is the year 2007.
• From 31 Dec 2007 to 31 Dec 2008 is the year 2008.
• From 31 Dec 2008 to 31 Dec 2009 is the year 2009.

Let's determine if these years are leap years and count the odd days:

• 2006: Not divisible by 4. This is a normal year. Number of odd days = 1.
• 2007: Not divisible by 4. This is a normal year. Number of odd days = 1.
• 2008: Divisible by 4. This is a leap year. Number of odd days = 2. (The leap day, Feb 29, is included in the period from 31 Dec 2007 to 31 Dec 2008).
• 2009: Not divisible by 4. This is a normal year. Number of odd days = 1.

Calculating Total Odd Days

The total number of odd days between 31 December 2005 and 31 December 2009 is the sum of the odd days for each full year in the interval:

Total odd days = (Odd days in 2006) + (Odd days in 2007) + (Odd days in 2008) + (Odd days in 2009)

Total odd days = \(1 + 1 + 2 + 1 = 5\) odd days.

Determining the Final Day of the Week

We started on a Saturday (31 December 2005). To find the day of the week for 31 December 2009, we move forward by the total number of odd days (5 days) from Saturday.

• Saturday + 1 day = Sunday
• Saturday + 2 days = Monday
• Saturday + 3 days = Tuesday
• Saturday + 4 days = Wednesday
• Saturday + 5 days = Thursday

Therefore, 31 December 2009 will be a Thursday.

Starting day: Saturday (corresponds to day 6 if Sunday is 0 or Saturday is 0 depending on convention, but we just add the days)

Add 5 odd days.

Saturday + 5 days = Thursday.

Revision Table: Calendar Concepts

Additional Information: Solving Day of the Week Problems

Calendar problems often appear in competitive exams. Mastering the concept of odd days is crucial. Here are some key points:

• The day of the week repeats every 7 days.
• Moving forward in time adds odd days to the day count. Moving backward subtracts odd days.
• The date moves forward by one day of the week for each normal year passed and by two days for each leap year passed. For example, if Jan 1, 2023, is a Sunday, Jan 1, 2024 (normal year 2023), will be a Monday. If Jan 1, 2024 (leap year), is a Monday, Jan 1, 2025, will be a Wednesday.
• Calculating the total number of days between two dates and finding the remainder when divided by 7 gives the total number of odd days.
• Be careful when counting odd days across centuries, as the 400-year rule for leap years (e.g., 1900 was not a leap year, but 2000 was) becomes important. In this problem, we stayed within one century where only the divisibility by 4 rule for non-century years and the divisibility by 400 rule for the year 2000 (which was a leap year but outside our range) matter. Only 2008 was relevant as a leap year here.