If the positive integer x is a multiple of 4 and the positive integer y is a multiple of 6,...

1. Translate the problem requirements: If x is a multiple of 4 and y is a multiple of 6, we need to determine which numbers xy MUST always be a multiple of, regardless of the specific values of x and y.
2. Express x and y in terms of their prime factorizations: Use the given constraints to write x and y in terms of their fundamental building blocks.
3. Determine the guaranteed prime factors of xy: Combine the factorizations to find what prime factors xy must always contain.
4. Test each option against our guaranteed factors: Check whether each of 8, 12, and 18 can be formed from the guaranteed prime factors of xy.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have two conditions:

• x is a positive integer that's a multiple of 4
• y is a positive integer that's a multiple of 6

We need to find what xy (the product of x and y) MUST always be a multiple of, no matter which specific values we choose for x and y.

The key word here is "must" - this means we're looking for numbers that xy will always be divisible by, regardless of our specific choices for x and y.

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

2. Express x and y in terms of their prime factorizations

Since x is a multiple of 4, we can write x in a simple way. Let's think about what 4 actually is:

\(4 = 2 \times 2 = \mathrm{2}^2\)

So if x is a multiple of 4, then x must contain at least two factors of 2. We can write: \(\mathrm{x} = 4\mathrm{k} = \mathrm{2}^2 \times \mathrm{k}\), where k is any positive integer.

Similarly, since y is a multiple of 6, let's break down 6:

\(6 = 2 \times 3\)

So if y is a multiple of 6, then y must contain at least one factor of 2 and one factor of 3. We can write: \(\mathrm{y} = 6\mathrm{j} = 2 \times 3 \times \mathrm{j}\), where j is any positive integer.

3. Determine the guaranteed prime factors of xy

Now let's find what prime factors xy must always have by multiplying our expressions:

\(\mathrm{xy} = (\mathrm{2}^2 \times \mathrm{k}) \times (2 \times 3 \times \mathrm{j})\)

\(\mathrm{xy} = \mathrm{2}^2 \times 2 \times 3 \times \mathrm{k} \times \mathrm{j}\)

\(\mathrm{xy} = \mathrm{2}^3 \times 3 \times \mathrm{k} \times \mathrm{j}\)

This tells us that xy must always contain at least three factors of 2 (that's \(\mathrm{2}^3 = 8\)) and at least one factor of 3.

So xy is guaranteed to be a multiple of \(\mathrm{2}^3 \times 3 = 8 \times 3 = 24\).

Process Skill: INFER - Drawing the non-obvious conclusion about guaranteed prime factors

4. Test each option against our guaranteed factors

Now let's check each option to see if it's guaranteed to divide xy:

Option I: Is xy always a multiple of 8?

Since xy contains \(\mathrm{2}^3 = 8\) as guaranteed factors, yes, xy must always be a multiple of 8.

Option II: Is xy always a multiple of 12?

Let's see what 12 requires: \(12 = 4 \times 3 = \mathrm{2}^2 \times 3\)

Since xy is guaranteed to have \(\mathrm{2}^3 \times 3\) (which includes \(\mathrm{2}^2 \times 3\)), yes, xy must always be a multiple of 12.

Option III: Is xy always a multiple of 18?

Let's check what 18 requires: \(18 = 2 \times 9 = 2 \times \mathrm{3}^2\)

For xy to be a multiple of 18, it needs at least one factor of 2 and two factors of 3. While xy is guaranteed to have \(\mathrm{2}^3\) (plenty of 2's), it's only guaranteed to have one factor of 3, not two. So xy is NOT always a multiple of 18.

Let's verify with an example: if \(\mathrm{x} = 4\) and \(\mathrm{y} = 6\), then \(\mathrm{xy} = 24\). And \(24 \div 18 = 1.33...\), so 24 is not divisible by 18.

Process Skill: CONSIDER ALL CASES - Testing each option systematically against our constraints

4. Final Answer

xy must always be a multiple of 8 (option I) and 12 (option II), but not necessarily 18 (option III).

Therefore, the answer is (B) I and II only.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "must be" vs "could be": Students often confuse what xy MUST always be a multiple of versus what xy COULD POSSIBLY be a multiple of. The question asks what xy must be a multiple of regardless of which specific values of x and y we choose, but students may pick values that work for some cases without checking if they work for ALL cases.

2. Not recognizing the need for prime factorization: Students may try to solve this by testing specific examples (like x=4, y=6) without realizing they need to find the guaranteed prime factors that will always be present in xy. This leads to incomplete analysis since examples alone cannot prove what must always be true.

3. Forgetting to consider the general form: Students may focus only on the smallest multiples (x=4, y=6) instead of recognizing that x could be any multiple of 4 (4, 8, 12, 16...) and y could be any multiple of 6 (6, 12, 18, 24...), missing the need to find what's guaranteed across all possibilities.

Errors while executing the approach

1. Incorrect prime factorization combination: When combining the prime factors from \(\mathrm{x} = \mathrm{2}^2\mathrm{k}\) and \(\mathrm{y} = 2 \cdot 3 \cdot \mathrm{j}\), students may incorrectly add exponents instead of multiplying them, getting \(\mathrm{xy} = \mathrm{2}^3 \cdot 3 \cdot \mathrm{k} \cdot \mathrm{j}\) wrong. They might write \(\mathrm{xy} = \mathrm{2}^2 \cdot 2 \cdot 3 = \mathrm{2}^4 \cdot 3\) instead of correctly identifying xy has \(\mathrm{2}^3 \cdot 3\) as guaranteed factors.

2. Confusing minimal guaranteed factors with possible factors: Students may correctly identify that xy contains \(\mathrm{2}^3 \cdot 3\) as guaranteed factors but then incorrectly conclude this means xy equals exactly 24, rather than understanding this means xy is always divisible by 24 but could have additional factors depending on k and j.

3. Arithmetic errors in checking divisibility: When testing if options like 8, 12, or 18 divide the guaranteed factors \(\mathrm{2}^3 \cdot 3 = 24\), students may make simple division errors or incorrectly break down the prime factorizations of the test numbers (like writing \(12 = 2 \cdot 6\) instead of \(12 = \mathrm{2}^2 \cdot 3\)).

Errors while selecting the answer

1. Incomplete verification of Option III: Students may correctly determine that Options I and II work, but fail to properly verify that Option III (\(18 = 2 \cdot \mathrm{3}^2\)) requires two factors of 3 while xy only guarantees one factor of 3. They might assume that since 24 is close to 18, it must work without checking the prime factorization requirement.

2. Selecting based on a single example: Even after doing the general analysis, students may second-guess themselves and revert to testing just one example (like x=4, y=6 giving xy=24), then incorrectly conclude the answer based on what works for that specific case rather than trusting their general proof.

Alternate Solutions

Smart Numbers Approach

Strategy: Choose specific values for x and y that satisfy the given conditions, then check which factors consistently appear in xy.

Step 1: Choose smart values for x and y
Since x must be a multiple of 4, let's try \(\mathrm{x} = 4\)
Since y must be a multiple of 6, let's try \(\mathrm{y} = 6\)

Step 2: Calculate xy
\(\mathrm{xy} = 4 \times 6 = 24\)

Step 3: Check which options divide 24
I. Does 8 divide 24? → \(24 \div 8 = 3\) ✓
II. Does 12 divide 24? → \(24 \div 12 = 2\) ✓
III. Does 18 divide 24? → \(24 \div 18 = 1.33...\) ✗

Step 4: Test with different smart values to confirm
Let's try \(\mathrm{x} = 8\), \(\mathrm{y} = 12\)
\(\mathrm{xy} = 8 \times 12 = 96\)
I. Does 8 divide 96? → \(96 \div 8 = 12\) ✓
II. Does 12 divide 96? → \(96 \div 12 = 8\) ✓
III. Does 18 divide 96? → \(96 \div 18 = 5.33...\) ✗

Step 5: Test one more pair to be certain
Let's try \(\mathrm{x} = 12\), \(\mathrm{y} = 6\)
\(\mathrm{xy} = 12 \times 6 = 72\)
I. Does 8 divide 72? → \(72 \div 8 = 9\) ✓
II. Does 12 divide 72? → \(72 \div 12 = 6\) ✓
III. Does 18 divide 72? → \(72 \div 18 = 4\) ✓

Step 6: Analyze the pattern
Wait - 18 divided 72 but not 24 or 96. This means 18 is NOT always a factor of xy.
However, 8 and 12 consistently divide xy in all our tests.

Conclusion: xy must always be a multiple of 8 and 12, but not necessarily 18.
Therefore, I and II only are correct.

Answer: (B) I and II only