If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does...

1. Translate the problem requirements: We need to find how many positive even divisors \(\mathrm{n}\) has (including \(\mathrm{n}\) itself), where \(\mathrm{n = 4p}\) and \(\mathrm{p}\) is a prime number greater than 2
2. Express n in terms of prime factorization: Since \(\mathrm{n = 4p}\) and \(\mathrm{4 = 2^2}\), we can write \(\mathrm{n = 2^2 \times p}\), which helps us identify all possible divisors
3. Systematically list all divisors of n: Using the prime factorization, identify all possible combinations of prime factors that form divisors
4. Filter for even divisors only: From our complete list of divisors, count only those that are even (contain at least one factor of 2)

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is asking us to find. We have a number \(\mathrm{n}\) that equals \(\mathrm{4p}\), where \(\mathrm{p}\) is a prime number greater than 2. We need to count how many positive even divisors \(\mathrm{n}\) has, including \(\mathrm{n}\) itself.

First, let's understand what a divisor is: A divisor of a number is any whole number that divides evenly into that number (with no remainder). For example, the divisors of 12 are 1, 2, 3, 4, 6, and 12.

Next, we need even divisors specifically - these are divisors that are divisible by 2. So from our example above, the even divisors of 12 would be 2, 4, 6, and 12.

Since \(\mathrm{p}\) is a prime greater than 2, we know \(\mathrm{p}\) must be odd (because the only even prime is 2, and \(\mathrm{p > 2}\)).

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

2. Express n in terms of prime factorization

Now let's rewrite \(\mathrm{n}\) using prime factorization to make finding divisors easier.

We know \(\mathrm{n = 4p}\). Since \(\mathrm{4 = 2^2}\), we can write:
\(\mathrm{n = 4p = 2^2 \times p}\)

This prime factorization tells us that \(\mathrm{n}\) contains exactly two factors of 2 and one factor of \(\mathrm{p}\) (where \(\mathrm{p}\) is an odd prime).

This representation is powerful because it shows us the "building blocks" we can use to construct all possible divisors.

3. Systematically list all divisors of n

Using our prime factorization \(\mathrm{n = 2^2 \times p}\), let's find all possible divisors by considering different combinations of the prime factors.

For any divisor of \(\mathrm{n}\), we can choose:

• 0, 1, or 2 factors of 2 (since we have \(\mathrm{2^2}\) available)
• 0 or 1 factor of \(\mathrm{p}\) (since we have \(\mathrm{p^1}\) available)

This gives us all possible divisors:

• \(\mathrm{2^0 \times p^0 = 1}\)
• \(\mathrm{2^1 \times p^0 = 2}\)
• \(\mathrm{2^2 \times p^0 = 4}\)
• \(\mathrm{2^0 \times p^1 = p}\)
• \(\mathrm{2^1 \times p^1 = 2p}\)
• \(\mathrm{2^2 \times p^1 = 4p = n}\)

So the complete list of divisors is: 1, 2, 4, \(\mathrm{p}\), \(\mathrm{2p}\), \(\mathrm{4p}\)

4. Filter for even divisors only

Now we need to identify which of these divisors are even. Remember, a number is even if it contains at least one factor of 2.

Looking at our list:

• \(\mathrm{1 = 2^0 \times p^0}\) → odd (no factors of 2)
• \(\mathrm{2 = 2^1 \times p^0}\) → even (contains factor of 2)
• \(\mathrm{4 = 2^2 \times p^0}\) → even (contains factors of 2)
• \(\mathrm{p = 2^0 \times p^1}\) → odd (\(\mathrm{p}\) is odd, no factors of 2)
• \(\mathrm{2p = 2^1 \times p^1}\) → even (contains factor of 2)
• \(\mathrm{4p = 2^2 \times p^1}\) → even (contains factors of 2)

Therefore, the even divisors are: 2, 4, \(\mathrm{2p}\), and \(\mathrm{4p}\)

Counting these up: we have exactly 4 positive even divisors.

Process Skill: APPLY CONSTRAINTS - Filtering systematically to meet the "even" requirement

5. Final Answer

We found that \(\mathrm{n = 4p}\) (where \(\mathrm{p}\) is a prime greater than 2) has exactly 4 positive even divisors: 2, 4, \(\mathrm{2p}\), and \(\mathrm{4p}\).

The answer is (C) Four.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint "p is a prime greater than 2"

Students may not realize that since \(\mathrm{p > 2}\) and \(\mathrm{p}\) is prime, \(\mathrm{p}\) must be odd. This is crucial because it affects which divisors will be even or odd. Without recognizing this, students might incorrectly assume \(\mathrm{p}\) could be even, leading to wrong divisor classifications.

2. Confusion about what "including n" means

The problem asks for positive even divisors "including \(\mathrm{n}\)". Students might misinterpret this and think they need to add \(\mathrm{n}\) as an extra divisor beyond their calculation, when actually \(\mathrm{n = 4p}\) is already one of the divisors they should find. This could lead to overcounting.

3. Not recognizing the need for prime factorization

Some students might try to find divisors by testing individual numbers rather than using the systematic prime factorization approach. This makes it much harder to ensure all divisors are found and correctly classified as even or odd.

Errors while executing the approach

1. Missing divisors when listing systematically

When using the prime factorization \(\mathrm{n = 2^2 \times p}\), students might forget that they need to consider all combinations of powers: \(\mathrm{2^0, 2^1, 2^2}\) combined with \(\mathrm{p^0, p^1}\). They might miss combinations like \(\mathrm{2^1 \times p^1 = 2p}\) or incorrectly think there are fewer total divisors.

2. Incorrectly identifying which divisors are even

Students might forget that a number is even if and only if it contains at least one factor of 2. They might incorrectly classify divisors like \(\mathrm{p}\) (which is odd) as even, or miss that divisors like \(\mathrm{2p}\) and \(\mathrm{4p}\) are definitely even because they contain factors of 2.

3. Arithmetic errors in expressing divisors

When writing out the divisors from the prime factorization, students might make simple errors like writing \(\mathrm{2^2 \times p^0 = 2}\) instead of 4, or confusing the notation and writing incorrect expressions for the divisors.

Errors while selecting the answer

1. Counting all divisors instead of just even divisors

After finding all 6 divisors (1, 2, 4, \(\mathrm{p}\), \(\mathrm{2p}\), \(\mathrm{4p}\)), students might count all of them and select answer choice (D) Six, forgetting that the question specifically asks only for even divisors. This is a classic case of not carefully re-reading what the question is asking for.

2. Forgetting to include n itself in the count

Students might correctly identify the even divisors 2, 4, and \(\mathrm{2p}\), but forget that \(\mathrm{n = 4p}\) is also an even divisor that should be included in their count. This would lead them to count only 3 even divisors and incorrectly select answer choice (B) Three.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a specific prime number greater than 2

Let's use \(\mathrm{p = 3}\) (the smallest prime greater than 2, making calculations simple)

Therefore: \(\mathrm{n = 4p = 4 \times 3 = 12}\)

Step 2: Express n in prime factorization form

\(\mathrm{n = 12 = 4 \times 3 = 2^2 \times 3}\)

Step 3: Find all positive divisors of n = 12

Using the prime factorization \(\mathrm{2^2 \times 3}\), the divisors are formed by taking:

• \(\mathrm{2^0 \times 3^0 = 1}\)
• \(\mathrm{2^1 \times 3^0 = 2}\)
• \(\mathrm{2^2 \times 3^0 = 4}\)
• \(\mathrm{2^0 \times 3^1 = 3}\)
• \(\mathrm{2^1 \times 3^1 = 6}\)
• \(\mathrm{2^2 \times 3^1 = 12}\)

All divisors: {1, 2, 3, 4, 6, 12}

Step 4: Identify the even divisors

Even divisors are those divisible by 2 (contain at least one factor of 2):

• 2 (even) ✓
• 4 (even) ✓
• 6 (even) ✓
• 12 (even) ✓

Step 5: Count the even divisors

The positive even divisors are: {2, 4, 6, 12}

Total count: 4 even divisors

Verification with another prime:

Let's verify with \(\mathrm{p = 5}\): \(\mathrm{n = 4 \times 5 = 20 = 2^2 \times 5}\)

All divisors: {1, 2, 4, 5, 10, 20}

Even divisors: {2, 4, 10, 20} → Again, 4 even divisors

This pattern will hold for any prime \(\mathrm{p > 2}\), confirming our answer of 4 even divisors.