Finding the Day of the Week Using Odd Days Method

This question asks us to find the day of the week on a specific date, 11 February 2006, given the day on another date, 11 August 2004, was a Wednesday. We can solve this type of problem by calculating the total number of 'odd days' between the two dates.

Understanding Odd Days

An 'odd day' is the remainder when the total number of days in a period is divided by 7 (since there are 7 days in a week). For example, if there are 10 days, $10 \div 7 = 1$ remainder 3, so there are 3 odd days. Adding the odd days to the starting day of the week gives the final day.

• 0 odd days means the same day.
• 1 odd day means the next day.
• 2 odd days means the day after next, and so on.

Step-by-Step Calculation

We need to calculate the number of days from 11 August 2004 to 11 February 2006 and find the total odd days.

Calculating Days from 11 August 2004 to 11 August 2005

This period is exactly one year. We need to check if this year includes a leap day (February 29th).

• The year 2004 is a leap year because it is divisible by 4.
• However, the period from 11 August 2004 to 11 August 2005 includes February 2005, which is not a leap year. February 2004 occurred *before* 11 August 2004.
• Therefore, the period from 11 August 2004 to 11 August 2005 has 365 days (a normal year).

Number of odd days in this period:

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

So, there is 1 odd day from 11 August 2004 to 11 August 2005.

Calculating Days from 11 August 2005 to 11 February 2006

We count the number of days month by month:

• August 2005: Remaining days = $31 - 11 = 20$ days
• September 2005: 30 days
• October 2005: 31 days
• November 2005: 30 days
• December 2005: 31 days
• January 2006: 31 days
• February 2006: 11 days (up to the 11th)

Total number of days in this period:

\( 20 + 30 + 31 + 30 + 31 + 31 + 11 = 184 \text{ days} \)

Number of odd days in this period:

\( \frac{184}{7} = 26 \text{ weeks and } 2 \text{ days remainder} \)

So, there are 2 odd days from 11 August 2005 to 11 February 2006.

Calculating Total Odd Days

Total odd days from 11 August 2004 to 11 February 2006 = (Odd days from 11 Aug 2004 to 11 Aug 2005) + (Odd days from 11 Aug 2005 to 11 Feb 2006)

Total odd days = $1 + 2 = 3$ odd days.

Determining the Final Day of the Week

The starting day is Wednesday. We need to move forward by 3 odd days.

• Wednesday + 1 day = Thursday
• Wednesday + 2 days = Friday
• Wednesday + 3 days = Saturday

Therefore, the day of the week on 11 February 2006 will be Saturday.

Total Odd Days = $1 + 2 = 3$

Starting Day = Wednesday

Final Day = Wednesday + 3 days = Saturday

Revision Table: Key Concepts in Calendar Problems

Additional Information on Calendar Calculations

Calendar problems often involve calculating the number of odd days over long periods, sometimes spanning centuries. Here are some helpful points:

• The number of odd days in 100 years is 5.
• The number of odd days in 200 years is $5 \times 2 = 10 \equiv 3 \pmod 7$. So, 3 odd days.
• The number of odd days in 300 years is $5 \times 3 = 15 \equiv 1 \pmod 7$. So, 1 odd day.
• The number of odd days in 400 years is $(5 \times 4) + 1$ (for the extra leap day in the 400th year) = $21 \equiv 0 \pmod 7$. So, 0 odd days.
• The cycle of odd days repeats every 400 years.
• To calculate the day of the week for a date from a fixed reference point (like 01/01/0001), you calculate total odd days from the reference point to the date.

This method of calculating odd days is fundamental to solving various calendar-related aptitude problems encountered in competitive exams.