What was the day of the week on 20th July 2009?
Monday
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.
The number of odd days in a period is the remainder when the total number of days in that period is divided by 7.
Let's calculate the odd days for a block of 400 years, as the calendar repeats every 400 years.
Odd days in 100 years:
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 |
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.
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.
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.
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 | \(3+0+3+2+3+2+6 = 19 \rightarrow 19 \pmod{7} = 5\) |
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.
Since we have a total of 1 odd day, the day of the week on 20th July 2009 was Monday.
Based on the calculation of odd days, the day of the week on 20th July 2009 was Monday.
| 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 | \(N \pmod{7}\) |
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:
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