Calculating the Day of the Week Between Two Dates

This problem requires us to determine the day of the week for a specific future date, given the day of the week for an earlier date. This can be solved using the concept of "odd days" in a calendar.

Understanding Odd Days and Leap Years

• An ordinary year has 365 days. When 365 is divided by 7 (days in a week), the remainder is 1. So, an ordinary year has 1 odd day.
• A leap year has 366 days. When 366 is divided by 7, the remainder is 2. So, a leap year has 2 odd days.
• Every 4th year is a leap year, except for century years that are not divisible by 400. Years like 1900, 2100 are not leap years, but 2000 is.

The difference in the day of the week between two dates depends on the total number of odd days accumulated during the period.

Step-by-Step Calculation

We need to find the day of the week for 29 November 1977, starting from 11 October 1969, which was a Saturday. We will calculate the number of odd days in the period from 11 October 1969 to 29 November 1977.

1. Odd Days Remaining in 1969 (from Oct 11 to Dec 31)

• Days in October remaining: 31 - 11 = 20 days
• Days in November: 30 days
• Days in December: 31 days
• Total days in this period = 20 + 30 + 31 = 81 days
• Odd days = $\frac{81}{7}$

When 81 is divided by 7: $81 = 11 \times 7 + 4$.

So, there are 4 odd days remaining in 1969 from October 11.

2. Odd Days in Full Years Between 1969 and 1977 (Years 1970 to 1976)

This period includes the years 1970, 1971, 1972, 1973, 1974, 1975, and 1976. This is a total of 7 full years.

• Identify Leap Years: Years divisible by 4 are leap years. In this period, 1972 and 1976 are leap years.
• Number of Leap Years = 2
• Number of Ordinary Years = Total years - Leap years = 7 - 2 = 5
• Odd days from Ordinary Years = $5 \times 1 = 5$ odd days
• Odd days from Leap Years = $2 \times 2 = 4$ odd days
• Total odd days in this period = $5 + 4 = 9$ odd days
• Equivalent odd days = $\frac{9}{7}$

When 9 is divided by 7: $9 = 1 \times 7 + 2$.

So, there are 2 odd days from the years 1970 to 1976.

3. Odd Days in 1977 (from Jan 1 to Nov 29)

We need to count the days from the beginning of 1977 until 29 November 1977. Note that 1977 is not a leap year, so February has 28 days.

• Jan: 31 days
• Feb: 28 days (1977 is ordinary)
• Mar: 31 days
• Apr: 30 days
• May: 31 days
• Jun: 30 days
• Jul: 31 days
• Aug: 31 days
• Sep: 30 days
• Oct: 31 days
• Nov: 29 days

Total days in 1977 until Nov 29 = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 29 = 333 days.

Odd days = $\frac{333}{7}$

When 333 is divided by 7: $333 = 47 \times 7 + 4$.

So, there are 4 odd days in 1977 until November 29.

4. Total Odd Days

Total odd days = Odd days from 1969 + Odd days from 1970-1976 + Odd days from 1977

Total odd days = 4 + 2 + 4 = 10 odd days.

Equivalent total odd days = $\frac{10}{7}$

When 10 is divided by 7: $10 = 1 \times 7 + 3$.

So, the net number of odd days from 11 October 1969 to 29 November 1977 is 3.

5. Determining the Final Day

The starting day is Saturday. We need to advance the day by the total number of odd days (3).

• Saturday + 1 day = Sunday
• Saturday + 2 days = Monday
• Saturday + 3 days = Tuesday

Therefore, the day of the week on 29 November 1977 will be Tuesday.

Conclusion

Based on the calculation of odd days, if 11 October 1969 was a Saturday, then 29 November 1977 will be a Tuesday.

Calendar Calculation Revision Table

Additional Information on Calendar Problems

Calendar problems are common in competitive exams and test reasoning ability. The key to solving these problems is a solid understanding of odd days and how they accumulate over time, especially considering leap years.

• Knowing the number of days in each month is essential.
• Understanding the rule for leap years (divisible by 4, except for century years not divisible by 400) is crucial.
• The days of the week repeat in a cycle of 7. This is why we use division by 7 to find odd days.
• Sunday is often considered the 0th or 7th day, Monday the 1st, Tuesday the 2nd, and so on, when working with odd days. In this case, we just advanced directly from the given day.