Calculating Day of the Week Across Years

This question asks us to find the day of the week for a specific date in 2008, given the day of the week for the same date in 1997. To solve this, we need to determine the total number of days between the two dates and then find the number of "odd days". Odd days are the remainder when the total number of days is divided by 7. Each odd day shifts the day of the week forward by one day.

Understanding the Time Period: 1997 to 2008

We are looking at the period from 12 October 1997 to 12 October 2008. This time span covers exactly 11 full years (2008 - 1997 = 11). To calculate the total number of days, we need to account for both ordinary years and leap years within this period.

Identifying Leap Years Between 1997 and 2008

A leap year occurs every 4 years, adding an extra day (February 29th) to the calendar. Years divisible by 100 are not leap years unless they are also divisible by 400.

We need to identify the leap years between 1997 and 2008 that include February 29th within our date range (Oct 12, 1997 to Oct 12, 2008).

• 1998: Not a leap year.
• 1999: Not a leap year.
• 2000: Divisible by 400, so it is a leap year. February 29, 2000 falls within the period.
• 2001: Not a leap year.
• 2002: Not a leap year.
• 2003: Not a leap year.
• 2004: Divisible by 4, so it is a leap year. February 29, 2004 falls within the period.
• 2005: Not a leap year.
• 2006: Not a leap year.
• 2007: Not a leap year.
• 2008: Divisible by 4, so it is a leap year. February 29, 2008 falls within the period (we are going up to Oct 12, 2008).

The leap years between 12 October 1997 and 12 October 2008 that contribute an extra day are 2000, 2004, and 2008. There are 3 leap years in this period.

Calculating Total Odd Days

An ordinary year has 365 days. When 365 is divided by 7:

\(365 \div 7 = 52 \text{ weeks and } 1 \text{ day remainder}\)

So, an ordinary year has 1 odd day.

A leap year has 366 days. When 366 is divided by 7:

\(366 \div 7 = 52 \text{ weeks and } 2 \text{ days remainder}\)

So, a leap year has 2 odd days.

In the 11-year period from 1997 to 2008, there are:

• Number of leap years = 3
• Number of ordinary years = Total years - Number of leap years = \(11 - 3 = 8\)

Now, we calculate the total number of odd days:

Total odd days = (Number of ordinary years \(\times\) Odd days in an ordinary year) + (Number of leap years \(\times\) Odd days in a leap year)

Total odd days = \((8 \times 1) + (3 \times 2)\)

Total odd days = \(8 + 6\)

Total odd days = \(14\)

To find the net shift in the day of the week, we find the remainder when the total odd days are divided by 7:

Net odd days = \(14 \div 7\)'s remainder

Net odd days = \(14 \bmod 7 = 0\)

There are 0 net odd days between 12 October 1997 and 12 October 2008.

Determining the Final Day

The starting day was Saturday. The number of net odd days is 0.

Final day = Starting day + Net odd days

Final day = Saturday + 0 days

Final day = Saturday

Therefore, if 12 October 1997 was a Saturday, then 12 October 2008 was also a Saturday.

Revision Table: Key Calendar Concepts

Additional Information on Day Calculation

Calculating days of the week for different dates relies on understanding the cycle of days (7 days in a week) and how this cycle is affected by the length of years (ordinary vs. leap). Each full week brings us back to the same day of the week. The "odd days" are what cause the day to change.

When calculating forward in time, each odd day moves the day one step forward (e.g., from Monday to Tuesday). When calculating backward in time, each odd day moves the day one step backward.

This method is useful for finding the day of the week for any date relative to a known date, provided you can correctly count the number of days and identify leap years in the interval.