June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note: 1984...

1. Translate the problem requirements: We need to find what day of the week June 25, 1987 falls on, given that June 25, 1982 was a Friday. We must account for the leap year 1984.
2. Calculate the total number of days between the two dates: Count the days from June 25, 1982 to June 25, 1987, being careful to include the extra day from the leap year 1984.
3. Apply the 7-day cycle principle: Since days of the week repeat every 7 days, find the remainder when the total days is divided by 7 to determine how many days forward from Friday we need to count.
4. Count forward from the starting day: Starting from Friday (June 25, 1982), count the remainder number of days forward to find the day of the week for June 25, 1987.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're looking for. We know that June 25, 1982 was a Friday, and we want to find out what day of the week June 25, 1987 was.

The key insight here is that days of the week follow a repeating pattern every 7 days. So if we can figure out how many days passed between these two dates, we can count forward from Friday to find our answer.

There's one important detail we must pay attention to: 1984 was a leap year, which means it had an extra day (February 29th). This will affect our total count.

Process Skill: TRANSLATE - Converting the calendar problem into a day-counting exercise

2. Calculate the total number of days between the two dates

Now let's count the days from June 25, 1982 to June 25, 1987. We're looking at exactly 5 years.

In a normal year, there are 365 days. So for 5 normal years, we would have:
\(5 \times 365 = 1825\) days

But we need to account for the leap year. The problem tells us that 1984 was a leap year, which means it had 366 days instead of 365 days. So we need to add 1 extra day:
\(1825 + 1 = 1826\) total days

Let's double-check this by thinking about it year by year:

• 1982 to 1983: 365 days
• 1983 to 1984: 365 days
• 1984 to 1985: 366 days (leap year)
• 1985 to 1986: 365 days
• 1986 to 1987: 365 days

Total: \(365 + 365 + 366 + 365 + 365 = 1826\) days ✓

3. Apply the 7-day cycle principle

Since days of the week repeat every 7 days, we don't need to count all 1,826 days forward. We just need to find the remainder when 1,826 is divided by 7.

Let's do this division:
\(1826 \div 7 = 260 \text{ remainder } 6\)

We can verify: \(260 \times 7 = 1820\), and \(1826 - 1820 = 6\)

This means that after 1,826 days, we'll be 6 days forward in the weekly cycle from where we started.

4. Count forward from the starting day

We started on Friday (June 25, 1982), and we need to count 6 days forward:

• Friday (starting day) + 1 day = Saturday
• Saturday + 1 day = Sunday
• Sunday + 1 day = Monday
• Monday + 1 day = Tuesday
• Tuesday + 1 day = Wednesday
• Wednesday + 1 day = Thursday

So June 25, 1987 fell on a Thursday.

Final Answer

June 25, 1987 fell on a Thursday.

Looking at our answer choices, this corresponds to choice E. Thursday.

We can verify our logic: we counted exactly 5 years with one leap year, calculated 1,826 total days, found that \(1826 \equiv 6 \pmod{7}\), and counted 6 days forward from Friday to get Thursday.

Common Faltering Points

Errors while devising the approach

Students often forget that leap years affect the day count calculation, or they may incorrectly assume that ALL years between 1982-1987 were leap years instead of recognizing that only 1984 was a leap year as stated in the problem.

Students may get confused about whether to count the starting date (June 25, 1982) or ending date (June 25, 1987) in their calculation, potentially adding or subtracting an extra day that throws off their final answer.

Students might attempt to literally count forward day by day for all 1,826 days instead of recognizing that they only need to find the remainder when dividing by 7, making the problem much more complex than necessary.

Errors while executing the approach

When calculating \(5 \times 365 + 1\), students may make simple multiplication or addition errors, getting 1,825 instead of 1,826, or miscalculating the total number of days across the 5-year period.

When dividing 1,826 by 7, students may incorrectly calculate the remainder as 5 instead of 6, or make errors in the division process (\(260 \times 7 = 1820\), remainder = 6).

Students may correctly recognize there's one leap year but incorrectly place it in the wrong year within the 1982-1987 timeframe, even though the problem clearly states 1984 was the leap year.

Errors while selecting the answer

When counting 6 days forward from Friday, students may miscount and land on Wednesday (counting only 5 steps) or Saturday (counting 7 steps), especially if they include or exclude the starting day incorrectly.

Students may correctly calculate the remainder as 6 but then count incorrectly, either starting their count from Thursday (one day before Friday) or making other sequential counting errors that lead them to Wednesday or Friday instead of Thursday.