If 21 March 2007 is a Wednesday, what would be the day of the week on 25 May 2015?
Monday
Finding the Day of the Week Between Dates
This question asks us to determine the day of the week on a specific date in the future, given the day of the week on an earlier date. We can solve this by calculating the total number of 'odd days' between the two dates. Odd days are the extra days remaining after forming complete weeks.
Steps to Calculate the Day of the Week
We need to find the day of the week on 25 May 2015, given that 21 March 2007 was a Wednesday. We will break this down into steps:
- Calculate the number of odd days between 21 March 2007 and 21 March 2015 (complete years).
- Calculate the number of odd days between 21 March 2015 and 25 May 2015 (days within the final year).
- Sum the odd days from both periods and find the final remainder when divided by 7.
- Add the final number of odd days to the day of the week of the starting date (Wednesday).
Step 1: Odd Days from 21 March 2007 to 21 March 2015
The period covers 8 full years from 2007 to 2015. The years are 2008, 2009, 2010, 2011, 2012, 2013, 2014, and 2015. We need to identify the leap years within this period. A leap year occurs every 4 years, except for years divisible by 100 but not by 400.
- Leap Years between 2007 and 2015 (inclusive of 2008, 2012 as they fall within the full years): 2008, 2012. There are 2 leap years.
- Normal Years: The remaining $8 - 2 = 6$ years are normal years.
Calculation of odd days for this 8-year period:
- A normal year has 365 days, which is 52 weeks and 1 day ($365 \div 7$ leaves a remainder of 1). So, a normal year has 1 odd day.
- A leap year has 366 days, which is 52 weeks and 2 days ($366 \div 7$ leaves a remainder of 2). So, a leap year has 2 odd days.
Total odd days for 8 years = (Number of normal years $\times$ 1 odd day) + (Number of leap years $\times$ 2 odd days)
Total odd days = $(6 \times 1) + (2 \times 2) = 6 + 4 = 10$ odd days.
To find the effective odd days, we find the remainder when 10 is divided by 7.
Effective odd days for the 8 years = $10 \div 7$ remainder = 3 odd days.
Step 2: Odd Days from 21 March 2015 to 25 May 2015
Now we calculate the number of days from 21 March 2015 to 25 May 2015.
- Days remaining in March 2015 (March has 31 days): $31 - 21 = 10$ days.
- Days in April 2015: 30 days.
- Days in May 2015 (up to 25th): 25 days.
Total number of days = $10 + 30 + 25 = 65$ days.
To find the number of odd days in this period, we divide the total days by 7 and find the remainder.
Odd days from 65 days = $65 \div 7$ remainder.
$65 = 9 \times 7 + 2$. The remainder is 2.
So, there are 2 odd days in this period.
Step 3: Total Odd Days
Total odd days from 21 March 2007 to 25 May 2015 is the sum of the odd days from Step 1 and Step 2.
Total odd days = (Odd days from years) + (Odd days from days within the year)
Total odd days = $3 + 2 = 5$ odd days.
Step 4: Determining the Final Day
The day of the week on 21 March 2007 was Wednesday. We need to move forward by the total number of odd days calculated in Step 3.
Day on 25 May 2015 = Wednesday + 5 days.
- Wednesday + 1 day = Thursday
- Wednesday + 2 days = Friday
- Wednesday + 3 days = Saturday
- Wednesday + 4 days = Sunday
- Wednesday + 5 days = Monday
Therefore, the day of the week on 25 May 2015 would be Monday.
Summary of Odd Day Calculation
| Period | Calculation | Odd Days |
|---|---|---|
| 21 March 2007 to 21 March 2015 (8 years) | 6 normal years $\times$ 1 + 2 leap years $\times$ 2 = 10 days. $10 \div 7$ remainder = 3 | 3 |
| 21 March 2015 to 25 May 2015 | March (10) + April (30) + May (25) = 65 days. $65 \div 7$ remainder = 2 | 2 |
| Total Odd Days | 3 + 2 = 5 | 5 |
Starting day: Wednesday
Final day: Wednesday + 5 days = Monday.
Revision Table: Calendar Concepts
| Concept | Description |
|---|---|
| Normal Year | 365 days, 52 weeks and 1 day. 1 odd day. |
| Leap Year | 366 days, 52 weeks and 2 days. 2 odd days. Occurs generally every 4 years (divisible by 4, except centenary years not divisible by 400). |
| Odd Days | The number of days remaining after dividing the total number of days in a period by 7. Used to find the shift in the day of the week. |
Additional Information: Day Calculation Tips
Understanding how to calculate odd days is key to solving calendar-based problems. Here are some extra tips:
- The day of the week repeats every 7 days. So, adding or subtracting multiples of 7 odd days brings you back to the same day.
- When calculating days between dates, be careful with the start and end dates. Including the end date but excluding the start date (or vice versa) is standard for counting intervals. In this case, we counted days *after* March 20, 2015, up to May 25, 2015.
- Counting odd days year by year simplifies the process for longer periods. Remember to account correctly for leap years.
- Months also have a fixed number of odd days:
- 31-day month: $31 \div 7$ remainder = 3 odd days.
- 30-day month: $30 \div 7$ remainder = 2 odd days.
- 28-day month (February, normal year): $28 \div 7$ remainder = 0 odd days.
- 29-day month (February, leap year): $29 \div 7$ remainder = 1 odd day.
By consistently applying the concept of odd days, you can solve various calendar problems accurately.
Important Questions from Clock and Calendar
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