A number n can be written as the product of four primer numbers, exactly two of which are the same....

1. Translate the problem requirements: Clarify what "product of four prime numbers, exactly two of which are the same" means in terms of prime factorization, and understand that we need to count all positive divisors including 1 and n itself.
2. Express n in standard prime factorization form: Convert the given description into the mathematical form \(\mathrm{n = p^a \times q^b \times r^c}\) to identify the exponents of each distinct prime factor.
3. Apply the divisor counting principle: Use the fundamental principle that for \(\mathrm{n = p^a \times q^b \times r^c}\), the number of divisors equals \(\mathrm{(a+1)(b+1)(c+1)}\) by considering all possible combinations of prime factor powers.
4. Calculate and verify the result: Compute the total number of divisors and confirm it matches one of the given answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is telling us in everyday language:

What does "product of four prime numbers" mean?
This means we multiply four prime numbers together to get our number n. For example, we might have something like \(\mathrm{2 \times 3 \times 5 \times 7}\).

What does "exactly two of which are the same" mean?
Out of these four prime numbers, exactly one prime appears twice, and the other primes appear once each. So instead of having four completely different primes, we have one prime that repeats once.

For example: \(\mathrm{2 \times 2 \times 3 \times 5}\) (here 2 appears twice)
Or: \(\mathrm{3 \times 3 \times 2 \times 7}\) (here 3 appears twice)

What are we looking for?
We need to count ALL the positive numbers that divide n evenly, including 1 and n itself. These are called divisors.

Process Skill: TRANSLATE - Converting the verbal description into a clear mathematical understanding

2. Express n in standard prime factorization form

Now let's convert our understanding into mathematical notation.

Since n is the product of four prime numbers where exactly two are the same, we can write:
\(\mathrm{n = p \times p \times q \times r}\)

Where p, q, and r are different prime numbers.

In standard mathematical form, this becomes:
\(\mathrm{n = p^2 \times q^1 \times r^1}\)

This tells us:

• Prime p appears with exponent 2
• Prime q appears with exponent 1
• Prime r appears with exponent 1

Let's use a concrete example to make this clear:
If \(\mathrm{n = 2 \times 2 \times 3 \times 5 = 60}\), then \(\mathrm{n = 2^2 \times 3^1 \times 5^1}\)

3. Apply the divisor counting principle

Here's the key insight: Every divisor of n is formed by choosing some power of each prime factor, up to the maximum power that appears in n.

Let's think about this step by step with our example \(\mathrm{n = 2^2 \times 3^1 \times 5^1}\):

For the prime 2 (which has exponent 2):
We can choose \(\mathrm{2^0 = 1}\), or \(\mathrm{2^1 = 2}\), or \(\mathrm{2^2 = 4}\)
That's 3 choices total (exponent + 1 = 2 + 1 = 3)

For the prime 3 (which has exponent 1):
We can choose \(\mathrm{3^0 = 1}\), or \(\mathrm{3^1 = 3}\)
That's 2 choices total (exponent + 1 = 1 + 1 = 2)

For the prime 5 (which has exponent 1):
We can choose \(\mathrm{5^0 = 1}\), or \(\mathrm{5^1 = 5}\)
That's 2 choices total (exponent + 1 = 1 + 1 = 2)

How many total combinations?
Since we make these choices independently for each prime, we multiply:
Total divisors = \(\mathrm{3 \times 2 \times 2 = 12}\)

In general form: For \(\mathrm{n = p^2 \times q^1 \times r^1}\), the number of divisors = \(\mathrm{(2+1) \times (1+1) \times (1+1) = 3 \times 2 \times 2 = 12}\)

Process Skill: APPLY CONSTRAINTS - Using the specific structure of the prime factorization

4. Calculate and verify the result

Let's verify our answer with the concrete example \(\mathrm{n = 2^2 \times 3^1 \times 5^1 = 60}\):

The divisors of 60 are:

• 1 (using \(\mathrm{2^0 \times 3^0 \times 5^0}\))
• 2 (using \(\mathrm{2^1 \times 3^0 \times 5^0}\))
• 3 (using \(\mathrm{2^0 \times 3^1 \times 5^0}\))
• 4 (using \(\mathrm{2^2 \times 3^0 \times 5^0}\))
• 5 (using \(\mathrm{2^0 \times 3^0 \times 5^1}\))
• 6 (using \(\mathrm{2^1 \times 3^1 \times 5^0}\))
• 10 (using \(\mathrm{2^1 \times 3^0 \times 5^1}\))
• 12 (using \(\mathrm{2^2 \times 3^1 \times 5^0}\))
• 15 (using \(\mathrm{2^0 \times 3^1 \times 5^1}\))
• 20 (using \(\mathrm{2^2 \times 3^0 \times 5^1}\))
• 30 (using \(\mathrm{2^1 \times 3^1 \times 5^1}\))
• 60 (using \(\mathrm{2^2 \times 3^1 \times 5^1}\))

Counting these up: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
That's exactly 12 divisors!

Final Answer

For any number n that can be written as the product of four prime numbers where exactly two are the same, n will have exactly 12 positive divisors.

The answer is D. 12.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "exactly two of which are the same"
Students often misunderstand this constraint in two ways:

• They think it means exactly two primes total (instead of four primes where one appears twice)
• They think it means two different primes each appear twice (like \(\mathrm{p^2 \times q^2}\) instead of \(\mathrm{p^2 \times q \times r}\))

This leads to wrong prime factorization forms like \(\mathrm{n = p \times q}\) or \(\mathrm{n = p^2 \times q^2}\), resulting in incorrect divisor counts.

2. Not recognizing the divisor counting formula
Many students don't know or forget the fundamental rule: for \(\mathrm{n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}}\), the number of divisors = \(\mathrm{(a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)}\). Instead, they try to manually list all divisors or use incorrect counting methods, which becomes very time-consuming and error-prone.

3. Assuming specific prime values
Some students get fixated on particular prime numbers (like always using 2, 3, 5) and think the answer depends on which specific primes are chosen. They don't realize that the divisor count depends only on the exponents in the prime factorization, not on the actual prime values themselves.

Errors while executing the approach

1. Incorrect exponent identification
Even when students understand the constraint correctly, they may write the wrong exponents in the prime factorization. For example, writing \(\mathrm{n = p^1 \times q^2 \times r^1}\) instead of \(\mathrm{n = p^2 \times q^1 \times r^1}\), which would give \(\mathrm{(1+1) \times (2+1) \times (1+1) = 12}\) instead of the correct \(\mathrm{(2+1) \times (1+1) \times (1+1) = 12}\). While this particular error still yields 12, similar mistakes could lead to wrong answers.

2. Arithmetic errors in multiplication
When applying the formula \(\mathrm{(2+1) \times (1+1) \times (1+1)}\), students might make simple calculation mistakes like getting \(\mathrm{3 \times 2 \times 2 = 10}\) instead of 12, or forgetting to add 1 to each exponent before multiplying.

Errors while selecting the answer

No likely faltering points - once students correctly calculate 12 divisors, the answer choice D is straightforward to select, and there are no additional conditions or transformations required.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a concrete example that satisfies the conditions

We need a number n that is the product of four prime numbers, exactly two of which are the same. Let's choose:

• The repeated prime: 2
• The two distinct primes: 3 and 5

So \(\mathrm{n = 2 \times 2 \times 3 \times 5 = 4 \times 3 \times 5 = 60}\)

In prime factorization form: \(\mathrm{n = 2^2 \times 3^1 \times 5^1}\)

Step 2: List all positive divisors of n = 60

A divisor of 60 must be of the form \(\mathrm{2^a \times 3^b \times 5^c}\) where:

• a can be 0, 1, or 2 (3 choices)
• b can be 0 or 1 (2 choices)
• c can be 0 or 1 (2 choices)

Let's systematically list all combinations:

When c = 0 (no factor of 5):

• \(\mathrm{2^0 \times 3^0 \times 5^0 = 1}\)
• \(\mathrm{2^1 \times 3^0 \times 5^0 = 2}\)
• \(\mathrm{2^2 \times 3^0 \times 5^0 = 4}\)
• \(\mathrm{2^0 \times 3^1 \times 5^0 = 3}\)
• \(\mathrm{2^1 \times 3^1 \times 5^0 = 6}\)
• \(\mathrm{2^2 \times 3^1 \times 5^0 = 12}\)

When c = 1 (includes factor of 5):

• \(\mathrm{2^0 \times 3^0 \times 5^1 = 5}\)
• \(\mathrm{2^1 \times 3^0 \times 5^1 = 10}\)
• \(\mathrm{2^2 \times 3^0 \times 5^1 = 20}\)
• \(\mathrm{2^0 \times 3^1 \times 5^1 = 15}\)
• \(\mathrm{2^1 \times 3^1 \times 5^1 = 30}\)
• \(\mathrm{2^2 \times 3^1 \times 5^1 = 60}\)

Step 3: Count the total divisors

Complete list of divisors: {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}

Total count: 12 divisors

Step 4: Verify this pattern holds for any valid n

Any number n satisfying the given conditions must have the form \(\mathrm{n = p^2 \times q \times r}\) where p, q, r are distinct primes. The number of divisors will always be \(\mathrm{(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12}\).

Therefore, the answer is D. 12