The fifteenth day of the month of June is a Saturday. What will be the last day of the next month which has 31 days?
Wednesday
This question asks us to determine the day of the week for a specific date in the future, given the day of the week for an earlier date. We are told that the fifteenth day of June is a Saturday. We need to find the day of the week for the last day of the next month which has 31 days.
The starting month is June. The next month is July. We need to confirm if July has 31 days. Yes, July is a month that has 31 days. The last day of July is the 31st. So, we need to find the day of the week for July 31st, given that June 15th is a Saturday.
To find the day of the week for July 31st, we need to calculate the total number of days from June 15th to July 31st. This calculation involves two parts:
June has 30 days. The number of days remaining in June from June 15th is the total days in June minus the day number we are starting from.
\( \text{Days remaining in June} = \text{Total days in June} - \text{Day in June we start from} \)
\( \text{Days remaining in June} = 30 - 15 = 15 \text{ days} \)
July is the month following June, and it has 31 days.
\( \text{Days in July} = 31 \text{ days} \)
The total number of days from June 15th to July 31st is the sum of the remaining days in June and the days in July.
\( \text{Total days} = \text{Days remaining in June} + \text{Days in July} \)
\( \text{Total days} = 15 + 31 = 46 \text{ days} \)
The day of the week repeats every 7 days. To find the day of the week after a certain number of days, we find the number of 'odd days'. Odd days are the remainder when the total number of days is divided by 7.
\( \text{Odd days} = \text{Total days} \pmod{7} \)
\( \text{Odd days} = 46 \pmod{7} \)
We divide 46 by 7:
\( 46 \div 7 = 6 \text{ with a remainder of } 4 \)
So, there are 4 odd days.
Starting from Saturday (June 15th), we move forward by the number of odd days calculated. The starting day is Saturday.
Therefore, 46 days after Saturday, the day will be Wednesday.
| Step | Description | Calculation | Result |
|---|---|---|---|
| 1 | Starting Date | June 15th (Saturday) | Saturday |
| 2 | Days remaining in June | 30 - 15 | 15 days |
| 3 | Days in July | 31 days | |
| 4 | Total days | 15 + 31 | 46 days |
| 5 | Odd days | \(46 \div 7\) (Remainder) | 4 odd days |
| 6 | Final Day | Saturday + 4 odd days | Wednesday |
The last day of the next month (July 31st) will be a Wednesday.
| Concept | Explanation |
|---|---|
| Odd Days | The remainder left after dividing the total number of days by 7. These represent the extra days beyond a complete week cycle. |
| Days in Months | The number of days each month has (Jan: 31, Feb: 28/29, Mar: 31, Apr: 30, May: 31, Jun: 30, Jul: 31, Aug: 31, Sep: 30, Oct: 31, Nov: 30, Dec: 31). |
| Leap Year | A year with 366 days, where February has 29 days. Occurs every 4 years, except for years divisible by 100 but not by 400. |
Calendar problems often rely on the concept of odd days because the sequence of days of the week is cyclical (repeats every 7 days). Knowing the number of odd days between two dates allows us to determine the day of the week for the later date if the day of the week for the earlier date is known.
To find the day of the week after \(N\) days from a given day:
In this specific problem, the total number of days from June 15th to July 31st was 46. The number of odd days was \(46 \pmod{7} = 4\). Starting from Saturday, moving forward 4 days gives us Sunday, Monday, Tuesday, Wednesday.
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