If 19 June 2001 is Tuesday, then what will be the day of the week on 21 August 2009?
Friday
Understanding Calendar Problems and Odd Days
Calendar problems often involve calculating the day of the week for a specific date, given the day of the week for another date. The key concept used to solve these problems is the 'odd day'.
An odd day is the number of days remaining after dividing the total number of days by 7 (since a week has 7 days). For example, 10 days have \(10 \pmod 7 = 3\) odd days. If today is Monday, after 10 days it will be Monday + 3 days = Thursday.
We need to find the day of the week on 21 August 2009, given that 19 June 2001 was a Tuesday. We will calculate the total number of odd days between these two dates.
Calculating Odd Days Step-by-Step
We can break down the time period from 19 June 2001 to 21 August 2009 into two parts:
- From 19 June 2001 to 19 June 2009 (covering full years).
- From 19 June 2009 to 21 August 2009 (within the final year).
Step 1: Calculate Odd Days from 19 June 2001 to 19 June 2009
This period covers exactly 8 years (2001-2002, 2002-2003, ..., 2008-2009). We need to count the number of ordinary years and leap years in this period.
- An ordinary year has 365 days, which is \(365 \pmod 7 = 1\) odd day.
- A leap year has 366 days, which is \(366 \pmod 7 = 2\) odd days.
The leap years between 2001 and 2009 are 2004 and 2008. Both these leap years include February 29th within the span from June to June.
- Number of leap years = 2 (2004, 2008)
- Number of ordinary years = Total years - Leap years = \(8 - 2 = 6\)
Total odd days in this period:
\[ (6 \text{ ordinary years} \times 1 \text{ odd day/year}) + (2 \text{ leap years} \times 2 \text{ odd days/year}) \] \[ = 6 + 4 = 10 \text{ odd days} \]The net odd days from 19 June 2001 to 19 June 2009 is \(10 \pmod 7 = 3\).
Step 2: Calculate Odd Days from 19 June 2009 to 21 August 2009
This calculation is within the year 2009, which is an ordinary year. We count the days from 19 June to 21 August.
- Days remaining in June: Total days in June (30) - 19 = 11 days. Odd days in June = \(11 \pmod 7 = 4\).
- Days in July: 31 days. Odd days in July = \(31 \pmod 7 = 3\).
- Days in August: Up to 21st = 21 days. Odd days in August = \(21 \pmod 7 = 0\).
Total odd days from 19 June 2009 to 21 August 2009:
\[ 11 \text{ days (June)} + 31 \text{ days (July)} + 21 \text{ days (August)} = 63 \text{ days} \]The net odd days in this period is \(63 \pmod 7 = 0\).
We can also calculate the odd days month by month:
\[ (\text{Odd days in June}) + (\text{Odd days in July}) + (\text{Odd days in August}) = 4 + 3 + 0 = 7 \text{ odd days} \] \[ 7 \pmod 7 = 0 \text{ odd days} \]Step 3: Calculate Total Odd Days
Total odd days from 19 June 2001 to 21 August 2009 is the sum of odd days from Step 1 and Step 2.
\[ \text{Total odd days} = (\text{Odd days across years}) + (\text{Odd days within year}) \] \[ \text{Total odd days} = 3 + 0 = 3 \]Step 4: Determine the Final Day of the Week
The day on 19 June 2001 was Tuesday. We need to move forward by the total number of odd days (3).
- Tuesday + 1 day = Wednesday
- Wednesday + 1 day = Thursday
- Thursday + 1 day = Friday
Therefore, the day of the week on 21 August 2009 will be Friday.
| Date Range | Calculation | Odd Days |
|---|---|---|
| 19 June 2001 to 19 June 2009 | 8 years (6 ordinary + 2 leap) | \(6 \times 1 + 2 \times 2 = 10 \equiv 3 \pmod 7\) |
| 19 June 2009 to 21 August 2009 | 11 days (June) + 31 days (July) + 21 days (August) = 63 days | \(63 \pmod 7 = 0\) |
| Total Odd Days | Sum of above | \(3 + 0 = 3\) |
Revision Table: Odd Days for Months (Ordinary Year)
Understanding the number of odd days in each month is helpful.
| Month | Number of Days | Odd Days (\(\text{Days} \pmod 7\)) |
|---|---|---|
| January | 31 | 3 |
| February (Ordinary) | 28 | 0 |
| February (Leap) | 29 | 1 |
| March | 31 | 3 |
| April | 30 | 2 |
| May | 31 | 3 |
| June | 30 | 2 |
| July | 31 | 3 |
| August | 31 | 3 |
| September | 30 | 2 |
| October | 31 | 3 |
| November | 30 | 2 |
| December | 31 | 3 |
Additional Information on Calendar Calculations
Here are some points to remember for calendar problems:
- A year is a leap year if it is divisible by 4, unless it is a century year not divisible by 400. For example, 2000 was a leap year (divisible by 400), but 1900 was not (divisible by 100 but not 400).
- The cycle of days of the week repeats every 7 days. Therefore, we use modulo 7 to find the odd days.
- When calculating odd days between two dates, sum up the odd days for the full years between the dates and then the odd days for the remaining months/days in the incomplete year span.
- If you calculate backward in time, subtract the odd days from the starting day. If calculating forward, add the odd days.
This method of calculating odd days is efficient for solving calendar-based reasoning questions often found in competitive exams.
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