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Question

What was the day of the week on 20th July 2009?

This question was previously asked in
SSC Stenographer 2019 Previous Year Paper (24-Dec-2020) (Shift 2)
The correct answer is

Monday

Calculating the Day of the Week for 20th July 2009

To find the day of the week for a specific date like 20th July 2009, we can use the concept of 'odd days'. Odd days are the number of days remaining after dividing the total number of days by 7 (since there are 7 days in a week). By calculating the total number of odd days from a known reference point up to the given date, we can determine the day.

Understanding Odd Days

The number of odd days in a period is the remainder when the total number of days in that period is divided by 7.

  • An ordinary year has 365 days: \(365 \div 7 = 52\) weeks and \(1\) day. So, an ordinary year has 1 odd day.
  • A leap year has 366 days: \(366 \div 7 = 52\) weeks and \(2\) days. So, a leap year has 2 odd days.

Calculating Odd Days for Centuries

Let's calculate the odd days for a block of 400 years, as the calendar repeats every 400 years.

Odd days in 100 years:

  • Number of leap years in 100 years (1-100 AD): \(100 \div 4 = 25\), but year 100 is not a leap year (divisible by 100 but not 400). So, \(24\) leap years.
  • Number of ordinary years in 100 years: \(100 - 24 = 76\).
  • Total odd days in 100 years = \((76 \times 1) + (24 \times 2) = 76 + 48 = 124\).
  • Odd days in 124 days = \(124 \div 7\) remainder = \(124 = 17 \times 7 + 5\). So, 5 odd days.

Odd days in 200 years = \(2 \times\) Odd days in 100 years = \(2 \times 5 = 10\) odd days. \(10 \div 7\) remainder = 3. So, 3 odd days in 200 years.

Odd days in 300 years = \(3 \times\) Odd days in 100 years = \(3 \times 5 = 15\) odd days. \(15 \div 7\) remainder = 1. So, 1 odd day in 300 years.

Odd days in 400 years = \(4 \times\) Odd days in 100 years + 1 odd day (for the leap year at 400 AD) = \((4 \times 5) + 1 = 21\) odd days. \(21 \div 7\) remainder = 0. So, 0 odd days in 400 years.

Period Odd Days
100 Years 5
200 Years 3
300 Years 1
400 Years 0

Calculating Odd Days up to the End of 2008

We need to calculate the total odd days from the beginning of the calendar (conventionally 1st Jan 1 AD) up to the day before 20th July 2009, which is 31st December 2008.

Total years completed by the end of 2008 is 2008 years.

We can break down 2008 years as follows:

\[2008 \text{ years} = 2000 \text{ years} + 8 \text{ years (2001 to 2008)}\]

Odd days in 2000 years:

\[2000 \text{ years} = 5 \times 400 \text{ years}\]

Since 400 years have 0 odd days, 2000 years also have \(5 \times 0 = 0\) odd days.

Odd days in the remaining 8 years (from 2001 to 2008):

We need to identify the number of ordinary and leap years in this period.

  • Years from 2001 to 2008 are: 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008.
  • Leap years in this period (divisible by 4, except year 2100, 2200, etc.): 2004, 2008. (There are 2 leap years).
  • Ordinary years in this period: \(8 - 2 = 6\).
  • Total odd days from 2001 to 2008 = \((6 \times 1) + (2 \times 2) = 6 + 4 = 10\).
  • Odd days in 10 days = \(10 \div 7\) remainder = 3.

Total odd days up to the end of 2008 = Odd days in 2000 years + Odd days in years 2001-2008 = \(0 + 3 = 3\) odd days.

Calculating Odd Days in 2009 up to 20th July

2009 is an ordinary year (not divisible by 4).

We need to calculate the odd days for each month from January 2009 up to 20th July 2009.

  • January (31 days): \(31 \div 7\) remainder = 3 odd days.
  • February (28 days in ordinary year): \(28 \div 7\) remainder = 0 odd days.
  • March (31 days): \(31 \div 7\) remainder = 3 odd days.
  • April (30 days): \(30 \div 7\) remainder = 2 odd days.
  • May (31 days): \(31 \div 7\) remainder = 3 odd days.
  • June (30 days): \(30 \div 7\) remainder = 2 odd days.
  • July (up to 20th): 20 days. \(20 \div 7\) remainder = 6 odd days.

Total odd days in 2009 up to 20th July = \(3 + 0 + 3 + 2 + 3 + 2 + 6 = 19\) odd days.

Odd days in 19 days = \(19 \div 7\) remainder = 5.

Month (2009) Number of Days Odd Days (\(\text{Days} \pmod{7}\))
January 31 3
February 28 0
March 31 3
April 30 2
May 31 3
June 30 2
July (up to 20th) 20 6
Total \(3+0+3+2+3+2+6 = 19 \rightarrow 19 \pmod{7} = 5\)

Determining the Day of the Week

Total odd days up to 20th July 2009 = Odd days up to end of 2008 + Odd days in 2009 up to 20th July

\[\text{Total Odd Days} = 3 + 5 = 8\]

Odd days in 8 days = \(8 \div 7\) remainder = 1.

Now we map the total odd days to the day of the week. A common mapping starts with 0 odd days corresponding to Sunday.

  • 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 a total of 1 odd day, the day of the week on 20th July 2009 was Monday.

Conclusion

Based on the calculation of odd days, the day of the week on 20th July 2009 was Monday.

Revision Table: Key Odd Day Values

Period Odd Days
Ordinary Year (365 days) 1
Leap Year (366 days) 2
100 Years 5
200 Years 3
300 Years 1
400 Years 0
Number of days N \(N \pmod{7}\)

Additional Information on Calendar Calculations

The method of calculating odd days is a standard technique for solving calendar-based reasoning problems. It relies on identifying the cycles in the calendar, primarily the 7-day week and the 400-year leap year cycle. While the calculation can seem lengthy, breaking it down into years, centuries, and months makes it manageable.

The definition of a leap year is crucial:

  • A year is a leap year if it is divisible by 4, EXCEPT for years divisible by 100 but not by 400.
  • Examples: 2000 was a leap year (divisible by 400). 1900 was not a leap year (divisible by 100 but not 400). 2004 was a leap year (divisible by 4). 2009 is not a leap year.
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