If p and s are different prime numbers, how many different positive factors does ps^2 have?

1. Translate the problem requirements: We need to find how many different positive factors the expression \(\mathrm{ps}^2\) has, where p and s are different prime numbers. A positive factor is a number that divides evenly into our expression.
2. Express the number in prime factorization form: Since p and s are already prime and different, \(\mathrm{ps}^2\) is already in its complete prime factorization form.
3. Apply the factor counting method: Use the systematic approach that for a number in the form \(\mathrm{a}^m \times \mathrm{b}^n\), the total factors equals \((\mathrm{m}+1)(\mathrm{n}+1)\).
4. Enumerate and verify the factors: List out all possible factors to confirm our calculation and ensure we haven't missed any.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is asking in everyday language. We have an expression \(\mathrm{ps}^2\), where p and s are different prime numbers. We need to find how many different positive factors this expression has.

A positive factor is simply a number that divides evenly into our expression \(\mathrm{ps}^2\) with no remainder. Think of it like this: if I can divide \(\mathrm{ps}^2\) by some number and get a whole number result, then that number is a factor.

For example, if p = 2 and s = 3, then \(\mathrm{ps}^2 = 2 \times 3^2 = 2 \times 9 = 18\). The factors of 18 would be all the numbers that divide evenly into 18.

Process Skill: TRANSLATE - Converting the mathematical expression into concrete understanding

2. Express the number in prime factorization form

Since p and s are already prime numbers and they're different from each other, our expression \(\mathrm{ps}^2\) is already in its prime factorization form. This is perfect because prime factorization is the most useful form for counting factors.

Let's think about what \(\mathrm{ps}^2\) looks like: it's the prime p raised to the power of 1, multiplied by the prime s raised to the power of 2. So we can write this as \(\mathrm{p}^1 \times \mathrm{s}^2\).

Using our concrete example: if p = 2 and s = 3, then \(\mathrm{ps}^2 = 2^1 \times 3^2 = 2 \times 9 = 18\).

3. Apply the factor counting method

Now here's the key insight: when we have a number expressed as prime factors with their powers, there's a systematic way to count all possible factors.

For any number in the form \(\mathrm{a}^m \times \mathrm{b}^n\) (where a and b are different primes), the total number of factors is \((\mathrm{m}+1) \times (\mathrm{n}+1)\). This works because:

- From the prime 'a', we can choose to include it 0, 1, 2, ..., or m times (that's m+1 choices)
- From the prime 'b', we can choose to include it 0, 1, 2, ..., or n times (that's n+1 choices)
- Every combination gives us a different factor

For our expression \(\mathrm{ps}^2 = \mathrm{p}^1 \times \mathrm{s}^2\):

- From prime p, we can choose it 0 or 1 times (that's 1+1 = 2 choices)
- From prime s, we can choose it 0, 1, or 2 times (that's 2+1 = 3 choices)
- Total factors = 2 × 3 = 6

4. Enumerate and verify the factors

Let's list out all the factors systematically to verify our answer:

1. Choose \(\mathrm{p}^0\) and \(\mathrm{s}^0\): This gives us 1
2. Choose \(\mathrm{p}^1\) and \(\mathrm{s}^0\): This gives us p
3. Choose \(\mathrm{p}^0\) and \(\mathrm{s}^1\): This gives us s
4. Choose \(\mathrm{p}^1\) and \(\mathrm{s}^1\): This gives us ps
5. Choose \(\mathrm{p}^0\) and \(\mathrm{s}^2\): This gives us \(\mathrm{s}^2\)
6. Choose \(\mathrm{p}^1\) and \(\mathrm{s}^2\): This gives us \(\mathrm{ps}^2\)

So the six different positive factors are: 1, p, s, ps, \(\mathrm{s}^2\), and \(\mathrm{ps}^2\).

Let's verify with our concrete example where p = 2 and s = 3:

1. 1
2. 2
3. 3
4. 6 (which is 2×3)
5. 9 (which is \(3^2\))
6. 18 (which is \(2 \times 3^2\))

We can check: 18 ÷ 1 = 18 ✓, 18 ÷ 2 = 9 ✓, 18 ÷ 3 = 6 ✓, 18 ÷ 6 = 3 ✓, 18 ÷ 9 = 2 ✓, 18 ÷ 18 = 1 ✓

Final Answer

The expression \(\mathrm{ps}^2\) has exactly 6 different positive factors. The answer is D. Six.

This result holds true regardless of which specific prime numbers p and s represent, as long as they are different primes.

Common Faltering Points

Errors while devising the approach

1. Missing the constraint that p and s are different primes

Students often overlook that the problem specifically states p and s are "different" prime numbers. They might treat this as a general case where p and s could be the same prime, which would change the entire structure of the problem. If p = s, then \(\mathrm{ps}^2\) would become \(\mathrm{p}^3\), leading to a completely different factor count of 4 instead of 6.

2. Confusing factors with prime factors

Some students misinterpret the question as asking for the number of prime factors rather than the total number of positive factors. This leads them to think the answer is just 2 (counting p and s), completely missing that we need to count ALL possible factors including 1, composite factors, and the number itself.

3. Not recognizing this as a factor-counting problem

Students might not immediately recognize that this requires the systematic factor-counting formula. Instead, they might try to approach it by picking specific values for p and s and manually listing factors, which works but is time-consuming and error-prone, especially under test conditions.

Errors while executing the approach

1. Incorrectly applying the factor counting formula

Even when students know the formula \((\mathrm{m}+1)(\mathrm{n}+1)\) for a number of the form \(\mathrm{p}^m \times \mathrm{s}^n\), they might misidentify the exponents. They could mistakenly think \(\mathrm{ps}^2\) has exponents (2,1) instead of (1,2), leading to the same numerical answer but showing conceptual confusion that could hurt them in more complex problems.

2. Forgetting to include 1 as a factor

When systematically listing factors, students sometimes forget that 1 is always a positive factor of any positive integer. This would lead them to count only 5 factors instead of 6, potentially selecting answer choice C instead of the correct answer D.

3. Double-counting or missing factors during enumeration

If students choose to verify by listing all factors explicitly, they might accidentally count a factor twice or miss one entirely. For example, they might list {1, p, s, ps, \(\mathrm{s}^2\)} and forget \(\mathrm{ps}^2\), or they might somehow count one factor twice in their enumeration.

Errors while selecting the answer

No likely faltering points - once students correctly execute the approach and find 6 factors, the answer selection is straightforward since 6 directly corresponds to choice D.

Alternate Solutions

Smart Numbers Approach

Strategy: Since the number of factors depends only on the exponents in prime factorization (not the specific prime values), we can use concrete prime numbers to find the pattern.

Step 1: Choose specific prime numbers
Let p = 2 and s = 3 (any two different primes work)

Step 2: Calculate \(\mathrm{ps}^2\)
\(\mathrm{ps}^2 = 2 \times 3^2 = 2 \times 9 = 18\)

Step 3: Find prime factorization
\(18 = 2^1 \times 3^2\)

Step 4: List all positive factors systematically
For \(2^1 \times 3^2\), each factor has the form \(2^a \times 3^b\) where:
• a can be 0 or 1 (2 choices)
• b can be 0, 1, or 2 (3 choices)

All factors:
• \(2^0 \times 3^0 = 1\)
• \(2^1 \times 3^0 = 2\)
• \(2^0 \times 3^1 = 3\)
• \(2^1 \times 3^1 = 6\)
• \(2^0 \times 3^2 = 9\)
• \(2^1 \times 3^2 = 18\)

Step 5: Count the factors
Total factors = 6

Verification: This matches the formula \((1+1)(2+1) = 2 \times 3 = 6\), confirming our answer works for any choice of different primes p and s.