On June 8, 2008, a computer set a speed record by completing arithmetic operations at a rate of approximately 1.026...

1. Translate the problem requirements: Convert the given rate (\(1.026 \times 10^{15}\) operations per second) to determine how long it takes to complete 1 million mega-operations, where 1 mega-operation = 1 million arithmetic operations. Express the final answer as a fraction of an hour.
2. Calculate total operations needed: Determine the total number of arithmetic operations required for 1 million mega-operations using the given definition.
3. Apply rate to find time in seconds: Use the computer's rate to calculate how many seconds are needed to complete the total operations.
4. Convert to fraction of an hour: Transform the time from seconds to hours by dividing by 3600, then compare with answer choices to find the closest match.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have a super-fast computer that can do arithmetic operations at a blazing rate of \(1.026 \times 10^{15}\) operations every second. That's over a quadrillion operations per second!

Our task is to figure out: if we want this computer to complete "1 million mega-operations," how long will that take as a fraction of an hour?

The key detail is that 1 mega-operation equals 1 million arithmetic operations. So we need to find the time for 1 million mega-operations and express it as a fraction of an hour.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Calculate total operations needed

Now let's figure out how many individual arithmetic operations we actually need.

If 1 mega-operation = 1 million arithmetic operations, then:
1 mega-operation = 1,000,000 arithmetic operations = \(10^6\) arithmetic operations

For 1 million mega-operations:
1 million mega-operations = 1,000,000 mega-operations = \(10^6\) mega-operations

Total arithmetic operations needed = \((10^6 \text{ mega-operations}) \times (10^6 \text{ arithmetic operations per mega-operation})\)
Total arithmetic operations = \(10^6 \times 10^6 = 10^{12}\) arithmetic operations

So we need the computer to perform 1 trillion individual arithmetic operations.

3. Apply rate to find time in seconds

Now we can use the computer's speed to find how long this will take.

The computer's rate is \(1.026 \times 10^{15}\) operations per second.

Using the basic relationship: Time = Work ÷ Rate

Time in seconds = (Total operations needed) ÷ (Operations per second)
Time in seconds = \((10^{12} \text{ operations}) ÷ (1.026 \times 10^{15} \text{ operations/second})\)

Time in seconds = \(10^{12} ÷ (1.026 \times 10^{15})\)
Time in seconds = \((1 ÷ 1.026) \times (10^{12} ÷ 10^{15})\)
Time in seconds = \((1 ÷ 1.026) \times 10^{-3}\)

Since \(1.026 ≈ 1\) (very close), we have:
Time in seconds ≈ \(1 \times 10^{-3} = 0.001\) seconds

4. Convert to fraction of an hour

Finally, we need to convert from seconds to hours. Since there are 3600 seconds in an hour:

Time in hours = Time in seconds ÷ 3600
Time in hours = \(10^{-3} ÷ 3600\)
Time in hours = \(10^{-3} ÷ (3.6 \times 10^3)\)
Time in hours = \((1 ÷ 3.6) \times (10^{-3} ÷ 10^3)\)
Time in hours = \((1 ÷ 3.6) \times 10^{-6}\)

Since \(1 ÷ 3.6 ≈ 0.28 ≈ 0.3\), we have:
Time in hours ≈ \(0.3 \times 10^{-6} = 3 \times 10^{-7}\)

Final Answer

The fraction of an hour needed is approximately \(3 \times 10^{-7}\).

Looking at our answer choices, this matches choice D: \(3 \times 10^{-7}\).

This makes sense - the computer is incredibly fast, so even 1 trillion operations take only a tiny fraction of an hour to complete.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what "1 million mega-operations" means: Students often get confused by the double conversion required here. They might think they need to find the time for just 1 mega-operation (\(10^6\) operations) instead of 1 million mega-operations (\(10^{12}\) total operations). This fundamental misunderstanding would lead them to calculate a much smaller amount of work needed.

2. Forgetting the final unit requirement: The question asks for the answer "as a fraction of an hour," but students might plan to solve for time in seconds and forget they need to convert to hours. This would cause them to stop their approach too early and select an answer that's off by a factor of 3600.

3. Not recognizing this as a basic rate problem: Some students get intimidated by the scientific notation and large numbers, missing that this is simply a Time = Work ÷ Rate problem. They might try to set up complex proportions or overthink the relationship between the given rate and required time.

Errors while executing the approach

1. Calculation errors with powers of 10: When computing \(10^6 \times 10^6 = 10^{12}\), students sometimes add the exponents incorrectly or get confused about whether to multiply or add. Similarly, when dividing \(10^{12}\) by \(10^{15}\), they might incorrectly calculate \(10^{-3}\) as \(10^3\) or make other exponent arithmetic errors.

2. Approximation errors with 1.026: Students might try to work with the exact value 1.026 throughout their calculations instead of recognizing that since \(1.026 ≈ 1\), they can simplify their arithmetic significantly. This leads to unnecessarily complex calculations and potential errors.

3. Unit conversion mistakes: When converting from seconds to hours by dividing by 3600, students often struggle with expressing 3600 in scientific notation (\(3.6 \times 10^3\)) and then performing the division correctly. They might also forget to convert altogether if they lost track of their planned approach.

Errors while selecting the answer

1. Choosing the wrong order of magnitude: Even if students perform most calculations correctly, they might select \(3 \times 10^{-6}\) or \(3 \times 10^{-8}\) instead of \(3 \times 10^{-7}\) due to small errors in their final approximation or unit conversion step. The scientific notation format makes it easy to be off by one order of magnitude.

2. Selecting an answer that doesn't make intuitive sense: Students might choose an answer like \(3 \times 10^{-1}\) (which represents 0.3 hours or about 18 minutes) without checking if this makes sense for a computer performing over a quadrillion operations per second. A quick reasonableness check would show this is far too long.