The question asks for the day of the week on 14 August 1966, given that 12 July 1965 was a Monday.
Step 1: Calculate Day Shift Over One Year
First, determine the day of the week for 12 July 1966.
- The period from 12 July 1965 to 12 July 1966 is exactly one year.
- The year 1966 is not a leap year, so it has 365 days.
- Calculate the day shift using modulo 7: $365 \pmod{7}$.
- $365 = (52 \times 7) + 1$. So, $365 \pmod{7} = 1$.
- This means 12 July 1966 was 1 day after Monday.
- Therefore, 12 July 1966 was a Tuesday.
Step 2: Calculate Days Remaining in the Target Period
Next, calculate the number of days between 12 July 1966 and 14 August 1966.
- Days remaining in July 1966: There are 31 days in July. Number of days = $31 - 12 = 19$ days.
- Days in August 1966: We need to count up to the 14th. Number of days = 14 days.
- Total number of days = $19 + 14 = 33$ days.
Step 3: Calculate the Day Shift for the Remaining Days
Calculate the shift in the day of the week for these 33 days.
- Use modulo 7: $33 \pmod{7}$.
- $33 = (4 \times 7) + 5$. So, $33 \pmod{7} = 5$.
- This means the day of the week advances by 5 days from the day on 12 July 1966.
Step 4: Determine the Final Day of the Week
Determine the day of the week for 14 August 1966.
- Start from the day calculated in Step 1: Tuesday.
- Add the shift calculated in Step 3: Tuesday + 5 days.
- Counting 5 days forward from Tuesday: Wednesday (1), Thursday (2), Friday (3), Saturday (4), Sunday (5).
Therefore, 14 August 1966 was a Sunday.
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