Understanding the Day of the Week Calculation

The question asks for the day of the week on a specific date: 15th September 2005. To solve this type of calendar problem, we use the concept of "odd days". Odd days are the extra days remaining after dividing the total number of days by 7 (since there are 7 days in a week).

The process involves calculating the total number of odd days from a reference point up to the given date. A common reference point considers 0 odd days as Sunday, 1 as Monday, and so on.

Calculating Odd Days

We need to calculate the total number of odd days up to 15th September 2005. We can break this down into:

1. Odd days in the years completed before 2005 (i.e., up to the end of 2004).
2. Odd days in the year 2005 up to 15th September.

Step 1: Odd days up to the end of 2004

We can calculate the odd days for centuries and then for the remaining years.

• Odd days in 100 years = 5
• Odd days in 200 years = $5 \times 2 = 10 \equiv 3$ odd days (since $10 \pmod 7 = 3$)
• Odd days in 300 years = $5 \times 3 = 15 \equiv 1$ odd day (since $15 \pmod 7 = 1$)
• Odd days in 400 years = $5 \times 4 + 1 = 21 \equiv 0$ odd days (since $21 \pmod 7 = 0$, the +1 is for the extra leap day in the 400th year).

Cycles of 400 years have 0 odd days. The year 2000 is a multiple of 400 (2000 = $5 \times 400$).

• Odd days up to the end of 2000 = 0 odd days.

Now, let's calculate odd days for the years from 2001 to 2004:

• A normal year has 365 days $\equiv 1$ odd day ($365 \pmod 7 = 1$).
• A leap year has 366 days $\equiv 2$ odd days ($366 \pmod 7 = 2$).

We need to check which years are leap years. A year is a leap year if it is divisible by 4, unless it is a century year not divisible by 400.

• 2001: Not divisible by 4 (Normal year) - 1 odd day.
• 2002: Not divisible by 4 (Normal year) - 1 odd day.
• 2003: Not divisible by 4 (Normal year) - 1 odd day.
• 2004: Divisible by 4 (Leap year) - 2 odd days.

Total odd days from 2001 to 2004 = $1 + 1 + 1 + 2 = 5$ odd days.

Total odd days up to the end of 2004 = (Odd days up to 2000) + (Odd days from 2001 to 2004) = $0 + 5 = 5$ odd days.

Step 2: Odd days in the year 2005 up to 15th September

We need to count the number of days from 1st January 2005 up to 15th September 2005 and find the number of odd days.

Days in each month in 2005 (2005 is not a leap year, so February has 28 days):

Total number of odd days in 2005 up to 15th September = $3 + 0 + 3 + 2 + 3 + 2 + 3 + 3 + 1 = 20$ odd days.

To find the final number of odd days from these 20 days: $20 \pmod 7 = 6$ odd days.

Step 3: Total Odd Days and Day of the Week

Total odd days up to 15th September 2005 = (Odd days up to end of 2004) + (Odd days in 2005 up to 15th Sep)

Total odd days = $5 + 6 = 11$ odd days.

Now, find the final number of odd days by taking the remainder when 11 is divided by 7:

$11 \pmod 7 = 4$ odd days.

We use the following mapping for the number of odd days to the day of the week:

• 0 odd days = Sunday
• 1 odd day = Monday
• 2 odd days = Tuesday
• 3 odd days = Wednesday
• 4 odd days = Thursday
• 5 odd days = Friday
• 6 odd days = Saturday

Since we have 4 odd days, the day of the week on 15th September 2005 was Thursday.

Summary of the Calendar Calculation

To find the day of the week for 15th September 2005:

1. Calculated odd days up to the end of the full years before 2005 (up to 2004). This resulted in 5 odd days.
2. Calculated odd days within the year 2005 up to the specific date (15th September). This resulted in 6 odd days.
3. Summed the odd days: $5 + 6 = 11$ odd days.
4. Found the remainder when the total odd days are divided by 7: $11 \pmod 7 = 4$ odd days.
5. Mapped the final number of odd days (4) to the corresponding day of the week, which is Thursday.

Therefore, the day of the week on 15th September 2005 was Thursday.

Revision Table: Calendar Concepts

Additional Information on Calendar Calculations

Calendar problems like finding the day of the week for a given date are common in reasoning and aptitude tests. The key is to efficiently calculate the total number of odd days. Remember the number of odd days in months (30-day months have 2 odd days, 31-day months have 3 odd days, February has 0 or 1 odd day depending on leap year) and for years (normal year 1, leap year 2).

Understanding the pattern of odd days over centuries ($5, 3, 1, 0$ for 100, 200, 300, 400 years respectively) is crucial for handling dates across different centuries.

Practice with different dates, including dates in different centuries and dates spanning February in a leap year, to become proficient in this method.