For any positive integer n, n > 1, the "length" of n is the number of positive primes (not necessary...

1. Translate the problem requirements: The "length" means counting all prime factors (including repeats) in the prime factorization. We need the highest possible count for any number less than 1000.
2. Identify the optimization strategy: To maximize the count of prime factors under a size constraint, use the smallest possible prime factor repeatedly.
3. Apply the smallest prime strategy: Use prime 2 as many times as possible while keeping the result under 1000.
4. Find the maximum power: Calculate the highest power of 2 that stays below 1000 to determine the maximum length.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what "length" means in this problem. The length of a number is simply how many prime factors we count when we break down the number completely into primes. For example, let's look at 50:

\(50 = 2 \times 25 = 2 \times 5 \times 5\)

When we count all the prime factors (including repeats), we get: 2, 5, 5. That's 3 prime factors total, so the length of 50 is 3.

Our goal is to find the largest possible length for any positive integer less than 1000. In other words, we want to find a number under 1000 that has the most prime factors when we count them all.

Process Skill: TRANSLATE - Converting the special definition of "length" into a clear mathematical concept

2. Identify the optimization strategy

Now let's think about this logically. If we want to maximize the number of prime factors while keeping our final number small (under 1000), what should we do?

Imagine you're trying to fit as many books as possible into a backpack with a weight limit. You'd choose the lightest books, right? Similarly, to get the most prime factors while staying under 1000, we should use the smallest possible prime factors.

The smallest prime number is 2. So our strategy is: use the prime 2 as many times as possible while keeping the result under 1000.

Process Skill: INFER - Drawing the non-obvious conclusion that smaller primes lead to more factors

3. Apply the smallest prime strategy

Let's use 2 repeatedly and see how far we can go:

\(2^1 = 2\) (length = 1)
\(2^2 = 4\) (length = 2)
\(2^3 = 8\) (length = 3)
\(2^4 = 16\) (length = 4)
\(2^5 = 32\) (length = 5)
\(2^6 = 64\) (length = 6)
\(2^7 = 128\) (length = 7)
\(2^8 = 256\) (length = 8)
\(2^9 = 512\) (length = 9)
\(2^{10} = 1024\) (length = 10, but this exceeds 1000!)

So \(2^9 = 512\) gives us a length of 9 and stays under our limit of 1000.

4. Find the maximum power

Let's verify our answer and make sure we can't do better:

• \(2^9 = 512 < 1000\) ✓ (length = 9)
• \(2^{10} = 1024 > 1000\) ✗ (would exceed our limit)

Could we use a different combination? Let's check: if we replace even one factor of 2 with a larger prime (like 3), we get a bigger number with the same or fewer factors:

• \(2^8 \times 3 = 256 \times 3 = 768\) (length = 9, but we could have used \(512 = 2^9\))
• \(2^7 \times 3^2 = 128 \times 9 = 1152 > 1000\) ✗

So using \(2^9 = 512\) gives us the maximum possible length of 9.

Process Skill: APPLY CONSTRAINTS - Ensuring our solution stays within the given limit

4. Final Answer

The greatest possible length of a positive integer less than 1000 is 9.

This is achieved by the number \(512 = 2^9\), which has 9 prime factors (all equal to 2).

The answer is B. 9

Common Faltering Points

Errors while devising the approach

Students may confuse "length" with the number of distinct prime factors rather than the total count of prime factors (including repetitions). For example, they might think \(50 = 2 \times 5^2\) has length 2 (counting only distinct primes 2 and 5) instead of length 3 (counting all prime factors: 2, 5, 5).

Students may not realize that to maximize the number of prime factors while staying under 1000, they should use the smallest possible prime (2) repeatedly. Instead, they might randomly try different combinations without a systematic strategy.

Students may think the number must be exactly 1000 or may forget that the number must be strictly less than 1000, leading them to consider \(2^{10} = 1024\) as a valid option.

Errors while executing the approach

Students may make arithmetic mistakes when computing successive powers of 2, such as calculating \(2^8 = 265\) instead of 256, or \(2^9 = 521\) instead of 512.

Students may calculate a few powers of 2 but stop before reaching \(2^9 = 512\), missing the maximum possible length. For instance, they might stop at \(2^7 = 128\) and conclude the answer is 7.

When checking if other prime combinations work better, students may make errors like calculating \(2^8 \times 3 = 256 \times 3 = 758\) and mistakenly think this gives a higher length than \(2^9 = 512\), forgetting that both have the same length of 9 factors.

Errors while selecting the answer

Students may select 512 (the actual number) instead of 9 (the length). However, since 512 is not among the answer choices, this would likely force them to reconsider.

Students who calculated \(2^{10} = 1024\) might select 10 as their answer, forgetting that 1024 exceeds the constraint of being less than 1000, so the maximum achievable length is actually 9.