Calculating Day of the Week: 22 July 1969 to 6 August 1985

This problem asks us to find the day of the week on a specific date (6 August 1985) given the day of the week on another date (22 July 1969). To solve this type of calendar problem, we need to calculate the total number of 'odd days' between the two dates. Odd days are the extra days remaining after forming complete weeks.

A normal year has 365 days, which is 52 weeks and 1 day ($365 = 52 \times 7 + 1$). So, a normal year has 1 odd day.

A leap year has 366 days, which is 52 weeks and 2 days ($366 = 52 \times 7 + 2$). So, a leap year has 2 odd days.

We need to calculate the total odd days from 22 July 1969 to 6 August 1985. We can break this period down:

1. Odd days from 22 July 1969 to the end of 1969.
2. Odd days for the full years from 1970 to 1984.
3. Odd days from the beginning of 1985 up to 6 August 1985.

Step-by-Step Day Calculation

Calculate Odd Days for 1969 (from July 22)

First, let's calculate the number of days remaining in 1969 starting from July 22:

• Days in July 1969 (from 22nd): 31 - 22 = 9 days
• Days in August 1969: 31 days
• Days in September 1969: 30 days
• Days in October 1969: 31 days
• Days in November 1969: 30 days
• Days in December 1969: 31 days

Total days in 1969 from July 22 = 9 + 31 + 30 + 31 + 30 + 31 = 162 days.

Number of odd days in this period = $162 \div 7$. The remainder is the number of odd days.

$162 = 7 \times 23 + 1$. So, the remainder is 1.

Odd days in 1969 (partial) = 1.

Calculate Odd Days for Full Years (1970-1984)

The period from 1970 to 1984 includes $1984 - 1970 + 1 = 15$ full years.

We need to identify the leap years within this period. A year is a leap year if it is divisible by 4 (except for years divisible by 100 but not by 400).

Leap years between 1970 and 1984 are: 1972, 1976, 1980, 1984.

Number of leap years = 4.

Number of normal years = Total years - Leap years = 15 - 4 = 11.

Odd days in normal years = 11 years $\times$ 1 odd day/year = 11 odd days.

Odd days in leap years = 4 years $\times$ 2 odd days/year = 8 odd days.

Total odd days from 1970 to 1984 = 11 + 8 = 19 odd days.

Number of net odd days for this period = $19 \div 7$. The remainder is the number of net odd days.

$19 = 7 \times 2 + 5$. So, the remainder is 5.

Odd days from 1970 to 1984 = 5.

Calculate Odd Days for 1985 (up to August 6)

Next, let's calculate the number of days from the beginning of 1985 up to 6 August 1985. Note that 1985 is not a leap year.

• Days in January 1985: 31
• Days in February 1985: 28 (not a leap year)
• Days in March 1985: 31
• Days in April 1985: 30
• Days in May 1985: 31
• Days in June 1985: 30
• Days in July 1985: 31
• Days in August 1985 (up to 6th): 6

Total days in 1985 up to August 6 = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 6 = 218 days.

Number of odd days in this period = $218 \div 7$. The remainder is the number of odd days.

$218 = 7 \times 31 + 1$. So, the remainder is 1.

Odd days in 1985 (partial) = 1.

Total Odd Days and Final Day Calculation

Now, we sum the odd days calculated for each period:

• Odd days in 1969 (partial): 1
• Odd days from 1970 to 1984: 5
• Odd days in 1985 (partial): 1

Total odd days = 1 + 5 + 1 = 7 odd days.

To find the net change in the day of the week, we find the remainder when the total odd days are divided by 7.

Net odd days = $7 \div 7$. The remainder is 0.

This means the day of the week will shift by 0 days from the starting day.

The starting day was Tuesday on 22 July 1969.

Adding the net odd days to the starting day: Tuesday + 0 days = Tuesday.

Therefore, the day of the week on 6 August 1985 will be Tuesday.

Let's summarize the odd days calculation:

Since the net odd days is 0, the day of the week on 6 August 1985 is the same as on 22 July 1969, which was Tuesday.

Revision Table: Key Concepts for Day Calculation

Additional Information: Calendar Calculations and Odd Days

Calendar problems often rely on understanding the concept of odd days. Each day of the week repeats every 7 days. By finding the total number of days between two dates and then calculating the remainder when divided by 7, we find out how many days forward (or backward, depending on the direction) the day of the week shifts.

To accurately calculate odd days over long periods, it is crucial to correctly identify leap years. Years divisible by 4 are generally leap years, except for centennial years (years divisible by 100) which are only leap years if they are also divisible by 400 (e.g., 1600, 2000 are leap years, but 1800, 1900 are not).

The number of odd days for a period determines the shift in the day of the week:

• 0 odd days: Same day
• 1 odd day: Shift forward by 1 day
• 2 odd days: Shift forward by 2 days
• 3 odd days: Shift forward by 3 days
• 4 odd days: Shift forward by 4 days
• 5 odd days: Shift forward by 5 days
• 6 odd days: Shift forward by 6 days

In our calculation from 22 July 1969 to 6 August 1985, we moved forward in time, so we added the odd days to the starting day. The total number of odd days was 7, which is equivalent to 0 net odd days ($7 \equiv 0 \pmod{7}$). Therefore, the day remained the same.