If is n is multiple of 5, and n = p^2 * q where p and q are prime, which...

1. Translate the problem requirements: We need to understand that \(\mathrm{n}\) is a multiple of 5, meaning 5 divides \(\mathrm{n}\) evenly. Since \(\mathrm{n = p^2 \times q}\) where \(\mathrm{p}\) and \(\mathrm{q}\) are prime numbers, we need to determine which answer choice expression must always be divisible by 25 (which is \(\mathrm{5^2}\)).
2. Analyze the prime factorization constraints: Since \(\mathrm{n}\) is a multiple of 5 and \(\mathrm{n = p^2 \times q}\) with \(\mathrm{p}\) and \(\mathrm{q}\) being prime, we need to determine which of \(\mathrm{p}\) or \(\mathrm{q}\) must be 5, and what this tells us about the overall structure.
3. Apply the divisibility requirement: For \(\mathrm{n}\) to be divisible by 5, and since 5 is prime, either \(\mathrm{p = 5}\) or \(\mathrm{q = 5}\) (since these are the only prime factors in \(\mathrm{n}\)).
4. Test each answer choice systematically: Examine what happens to each expression when either \(\mathrm{p = 5}\) or \(\mathrm{q = 5}\), checking which one is guaranteed to be divisible by 25 in all valid cases.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English. We have a number \(\mathrm{n}\) that is a multiple of 5, which means when we divide \(\mathrm{n}\) by 5, we get a whole number with no remainder. We're also told that \(\mathrm{n}\) can be written as \(\mathrm{p^2 \times q}\), where both \(\mathrm{p}\) and \(\mathrm{q}\) are prime numbers (numbers that can only be divided by 1 and themselves, like 2, 3, 5, 7, 11, etc.).

Now we need to find which of the given expressions must always be a multiple of 25. Remember that \(\mathrm{25 = 5 \times 5 = 5^2}\), so for something to be a multiple of 25, it needs to have at least two factors of 5 in its prime factorization.

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

2. Analyze the prime factorization constraints

Since \(\mathrm{n}\) is a multiple of 5, and \(\mathrm{n = p^2 \times q}\) where \(\mathrm{p}\) and \(\mathrm{q}\) are prime, let's think about what this means. The number 5 is itself prime, so for \(\mathrm{n}\) to be divisible by 5, either \(\mathrm{p}\) must equal 5 or \(\mathrm{q}\) must equal 5 (or both, but that's not possible since \(\mathrm{p}\) and \(\mathrm{q}\) are different primes in this context).

Let's consider both cases:

• Case 1: \(\mathrm{p = 5}\), so \(\mathrm{n = 5^2 \times q = 25q}\)
• Case 2: \(\mathrm{q = 5}\), so \(\mathrm{n = p^2 \times 5}\)

In Case 1, \(\mathrm{n}\) already contains 25 as a factor. In Case 2, \(\mathrm{n}\) contains only one factor of 5.

Process Skill: INFER - Drawing the non-obvious conclusion about which prime must be 5

3. Apply the divisibility requirement

Now let's think about what we need. We want an expression that is ALWAYS a multiple of 25, regardless of which case we're in.

Let's test with concrete examples to make this clear:

• Example 1: If \(\mathrm{p = 5}\) and \(\mathrm{q = 3}\), then \(\mathrm{n = 25 \times 3 = 75}\)
• Example 2: If \(\mathrm{p = 3}\) and \(\mathrm{q = 5}\), then \(\mathrm{n = 9 \times 5 = 45}\)

Both examples give us valid values of \(\mathrm{n}\) that are multiples of 5. Now we need to check which answer choice works for both scenarios.

4. Test each answer choice systematically

Let's check each option using our examples:

Choice A: \(\mathrm{p^2}\)

• Example 1: \(\mathrm{p^2 = 5^2 = 25}\) ✓ (multiple of 25)
• Example 2: \(\mathrm{p^2 = 3^2 = 9}\) ✗ (not a multiple of 25)

Choice B: \(\mathrm{q^2}\)

• Example 1: \(\mathrm{q^2 = 3^2 = 9}\) ✗ (not a multiple of 25)
• Example 2: \(\mathrm{q^2 = 5^2 = 25}\) ✓ (multiple of 25)

Choice C: \(\mathrm{pq}\)

• Example 1: \(\mathrm{pq = 5 \times 3 = 15}\) ✗ (not a multiple of 25)
• Example 2: \(\mathrm{pq = 3 \times 5 = 15}\) ✗ (not a multiple of 25)

Choice D: \(\mathrm{p^2q^2}\)

• Example 1: \(\mathrm{p^2q^2 = 25 \times 9 = 225 = 9 \times 25}\) ✓ (multiple of 25)
• Example 2: \(\mathrm{p^2q^2 = 9 \times 25 = 225 = 9 \times 25}\) ✓ (multiple of 25)

Choice E: \(\mathrm{p^3q}\)

• Example 1: \(\mathrm{p^3q = 125 \times 3 = 375 = 15 \times 25}\) ✓ (multiple of 25)
• Example 2: \(\mathrm{p^3q = 27 \times 5 = 135}\) ✗ (not a multiple of 25)

Only Choice D works in both cases! This makes sense because \(\mathrm{p^2q^2}\) includes both \(\mathrm{p^2}\) and \(\mathrm{q^2}\), so regardless of which prime equals 5, we're guaranteed to have at least \(\mathrm{5^2 = 25}\) as a factor.

Process Skill: CONSIDER ALL CASES - Systematically checking both possible scenarios to ensure our answer works universally

Final Answer

The answer is D. \(\mathrm{p^2q^2}\)

This expression must always be a multiple of 25 because it contains both \(\mathrm{p^2}\) and \(\mathrm{q^2}\) as factors. Since either \(\mathrm{p = 5}\) or \(\mathrm{q = 5}\) (one of them must be 5 for \(\mathrm{n}\) to be a multiple of 5), the expression \(\mathrm{p^2q^2}\) will always contain at least \(\mathrm{5^2 = 25}\) as a factor, making it a multiple of 25 in all valid cases.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint "n is a multiple of 5"
Students often overlook that since \(\mathrm{n = p^2\times q}\) where \(\mathrm{p}\) and \(\mathrm{q}\) are prime, and \(\mathrm{n}\) must be divisible by 5, then either \(\mathrm{p}\) or \(\mathrm{q}\) must equal 5 (since 5 is prime). They might think any combination of primes works without considering this critical constraint.

2. Confusing "multiple of 25" with "multiple of 5"
Students may not realize that being a multiple of 25 requires having \(\mathrm{5^2 = 25}\) as a factor, meaning two factors of 5 are needed in the prime factorization. They might incorrectly think one factor of 5 is sufficient.

3. Not considering all possible cases
Students might only consider one scenario (like \(\mathrm{p = 5}\)) and forget to check that their answer must work when \(\mathrm{q = 5}\) as well. This leads to choosing answers that work in some cases but not all.

Errors while executing the approach

1. Testing with insufficient examples
Students might test answer choices with only one example instead of checking both cases (\(\mathrm{p = 5}\) and \(\mathrm{q = 5}\)). This incomplete testing can lead them to incorrectly validate wrong answer choices.

2. Arithmetic errors when calculating powers
When computing expressions like \(\mathrm{p^2q^2}\), students may make calculation errors, especially when substituting \(\mathrm{p = 5}\) or \(\mathrm{q = 5}\). For example, miscalculating \(\mathrm{5^2}\) as 10 instead of 25.

3. Incorrectly identifying factors of 25
Students may incorrectly determine whether a number is a multiple of 25. For instance, they might think 135 (from \(\mathrm{p^3q = 27\times 5}\)) is divisible by 25 due to hasty mental math.

Errors while selecting the answer

1. Choosing the first option that works in one case
Students might select answer choice A (\(\mathrm{p^2}\)) because it equals 25 when \(\mathrm{p = 5}\), without verifying that it fails when \(\mathrm{p \neq 5}\). They stop checking once they find one case where it works.

2. Selecting based on partial pattern recognition
Students might choose answer choice E (\(\mathrm{p^3q}\)) because it works when \(\mathrm{p = 5}\), thinking that higher powers are "more likely" to be multiples of 25, without completing the verification for all cases.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose smart values that satisfy the constraints

Since \(\mathrm{n}\) is a multiple of 5 and \(\mathrm{n = p^2 \times q}\) where \(\mathrm{p}\) and \(\mathrm{q}\) are prime, we need the factor 5 to appear in \(\mathrm{n}\). This means either \(\mathrm{p = 5}\) or \(\mathrm{q = 5}\) (or both, but let's test the simpler cases first).

Let's test two strategic cases:

Case 1: Let \(\mathrm{p = 5}\) and \(\mathrm{q = 2}\) (both prime)
Then \(\mathrm{n = 5^2 \times 2 = 25 \times 2 = 50}\) ✓ (multiple of 5)

Case 2: Let \(\mathrm{p = 2}\) and \(\mathrm{q = 5}\) (both prime)
Then \(\mathrm{n = 2^2 \times 5 = 4 \times 5 = 20}\) ✓ (multiple of 5)

Step 2: Test each answer choice with our smart numbers

For Case 1 (\(\mathrm{p = 5, q = 2}\)):

1. \(\mathrm{p^2 = 5^2 = 25}\) ✓ (multiple of 25)
2. \(\mathrm{q^2 = 2^2 = 4}\) ✗ (not multiple of 25)
3. \(\mathrm{pq = 5 \times 2 = 10}\) ✗ (not multiple of 25)
4. \(\mathrm{p^2q^2 = 5^2 \times 2^2 = 25 \times 4 = 100}\) ✓ (multiple of 25)
5. \(\mathrm{p^3q = 5^3 \times 2 = 125 \times 2 = 250}\) ✓ (multiple of 25)

For Case 2 (\(\mathrm{p = 2, q = 5}\)):

1. \(\mathrm{p^2 = 2^2 = 4}\) ✗ (not multiple of 25)
2. \(\mathrm{q^2 = 5^2 = 25}\) ✓ (multiple of 25)
3. \(\mathrm{pq = 2 \times 5 = 10}\) ✗ (not multiple of 25)
4. \(\mathrm{p^2q^2 = 2^2 \times 5^2 = 4 \times 25 = 100}\) ✓ (multiple of 25)
5. \(\mathrm{p^3q = 2^3 \times 5 = 8 \times 5 = 40}\) ✗ (not multiple of 25)

Step 3: Identify the pattern

Looking at our results:

• Choice A works only when \(\mathrm{p = 5}\)
• Choice B works only when \(\mathrm{q = 5}\)
• Choice C never works
• Choice D works in BOTH cases
• Choice E works only when \(\mathrm{p = 5}\)

Step 4: Verify the logic

Choice D = \(\mathrm{p^2q^2}\) always works because:

• When \(\mathrm{p = 5}\): \(\mathrm{p^2q^2}\) contains \(\mathrm{5^2 = 25}\) as a factor
• When \(\mathrm{q = 5}\): \(\mathrm{p^2q^2}\) contains \(\mathrm{5^2 = 25}\) as a factor

Since exactly one of \(\mathrm{p}\) or \(\mathrm{q}\) must equal 5 (for \(\mathrm{n}\) to be divisible by 5), \(\mathrm{p^2q^2}\) will always contain \(\mathrm{5^2 = 25}\) as a factor, making it always divisible by 25.

Answer: D