Which of the following CANNOT be the least common multiple of two positive integers x and y?

1. Translate the problem requirements: We need to identify which expression among the five choices can NEVER be the least common multiple (LCM) of two positive integers x and y. The LCM is the smallest positive integer that both x and y divide into evenly.
2. Test each option with concrete examples: For each answer choice, try to find specific values of x and y where that expression equals their LCM, or determine if it's impossible.
3. Apply LCM properties to eliminate impossible cases: Use the fundamental property that \(\mathrm{LCM(x,y)}\) must be greater than or equal to both x and y, and analyze which expressions can violate this requirement.
4. Identify the impossible option: Determine which expression can become negative or fail to satisfy the basic requirements of being an LCM.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in plain English. The LCM (least common multiple) of two numbers is the smallest positive number that both original numbers can divide into evenly.

For example, if we have numbers 4 and 6:

• Multiples of 4: 4, 8, 12, 16, 20, 24, ...
• Multiples of 6: 6, 12, 18, 24, 30, ...
• The smallest number that appears in both lists is 12, so \(\mathrm{LCM(4,6) = 12}\)

The question asks which expression CANNOT be the LCM of two positive integers x and y. This means we need to find which answer choice could never equal \(\mathrm{LCM(x,y)}\) for any positive values of x and y.

Process Skill: TRANSLATE - Converting the problem from abstract mathematical language to concrete understanding

2. Test each option with concrete examples

Let's examine each answer choice to see if we can find values of x and y where that expression equals their LCM:

Choice (A): \(\mathrm{xy}\)
Can \(\mathrm{LCM(x,y) = xy}\)? Let's try \(\mathrm{x = 3, y = 5}\)

• These are relatively prime (no common factors except 1)
• \(\mathrm{LCM(3,5) = 15 = 3 × 5 = xy}\) ✓

Choice (B): \(\mathrm{x}\)
Can \(\mathrm{LCM(x,y) = x}\)? Let's try \(\mathrm{x = 6, y = 3}\)

• Since 3 divides evenly into 6, the \(\mathrm{LCM(6,3) = 6 = x}\) ✓

Choice (C): \(\mathrm{y}\)
Can \(\mathrm{LCM(x,y) = y}\)? Let's try \(\mathrm{x = 4, y = 12}\)

• Since 4 divides evenly into 12, the \(\mathrm{LCM(4,12) = 12 = y}\) ✓

Choice (D): \(\mathrm{x - y}\)
Can \(\mathrm{LCM(x,y) = x - y}\)? Let's think about this...
If \(\mathrm{x = 5}\) and \(\mathrm{y = 3}\), then \(\mathrm{x - y = 2}\)

• \(\mathrm{LCM(5,3) = 15}\), but \(\mathrm{x - y = 2}\)
• For the LCM to equal \(\mathrm{x - y}\), we'd need \(\mathrm{LCM(5,3) = 2}\)
• But 2 is smaller than both 5 and 3, which violates a fundamental property

Choice (E): \(\mathrm{x + y}\)
Can \(\mathrm{LCM(x,y) = x + y}\)? Let's try \(\mathrm{x = 2, y = 3}\)

• \(\mathrm{LCM(2,3) = 6 = 2 + 3 = x + y}\) ✓

3. Apply LCM properties to eliminate impossible cases

Here's the key insight: The LCM of any two positive integers must be greater than or equal to each of the original numbers.

In mathematical terms: \(\mathrm{LCM(x,y) ≥ x}\) and \(\mathrm{LCM(x,y) ≥ y}\)

This makes sense because if a number is a multiple of x, it must be at least as large as x itself.

Now let's check if \(\mathrm{x - y}\) can satisfy this requirement:

• For \(\mathrm{LCM(x,y) = x - y}\) to work, we need: \(\mathrm{x - y ≥ x}\) and \(\mathrm{x - y ≥ y}\)
• The first condition gives us: \(\mathrm{x - y ≥ x}\), which simplifies to \(\mathrm{-y ≥ 0}\), or \(\mathrm{y ≤ 0}\)
• But the problem states that x and y are positive integers!
• This creates a contradiction.

Process Skill: INFER - Drawing the non-obvious conclusion that the LCM size constraint eliminates option D

4. Identify the impossible option

Let's verify our reasoning with a specific example:

Take any positive integers where \(\mathrm{x > y}\), say \(\mathrm{x = 7, y = 3}\):

• \(\mathrm{x - y = 7 - 3 = 4}\)
• But \(\mathrm{LCM(7,3) = 21}\) (since 7 and 3 are relatively prime)
• We would need \(\mathrm{LCM(7,3) = 4}\), but 4 cannot be a multiple of 7

Take any positive integers where \(\mathrm{x < y}\), say \(\mathrm{x = 3, y = 7}\):

• \(\mathrm{x - y = 3 - 7 = -4}\)
• The LCM must be positive, so it cannot equal -4

Take \(\mathrm{x = y}\), say \(\mathrm{x = 5, y = 5}\):

• \(\mathrm{x - y = 0}\)
• \(\mathrm{LCM(5,5) = 5}\), but we would need it to equal 0
• Zero cannot be an LCM since LCMs are positive

Therefore, \(\mathrm{x - y}\) can NEVER be the LCM of two positive integers.

5. Final Answer

The answer is (D) \(\mathrm{x - y}\).

The expression \(\mathrm{x - y}\) cannot be the least common multiple of two positive integers x and y because:

1. When \(\mathrm{x > y}\): \(\mathrm{x - y}\) is smaller than x, violating the requirement that \(\mathrm{LCM(x,y) ≥ x}\)
2. When \(\mathrm{x < y}\): \(\mathrm{x - y}\) is negative, but LCMs must be positive
3. When \(\mathrm{x = y}\): \(\mathrm{x - y = 0}\), but LCMs must be positive

All other choices can be LCMs under appropriate conditions, as demonstrated with concrete examples.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "CANNOT be" means
Students often confuse this with "what IS the LCM" and start looking for what the LCM actually equals, rather than identifying what it can never equal. This leads them to pick an option that CAN be an LCM instead of one that CANNOT be.

2. Forgetting the fundamental constraint that LCM ≥ both original numbers
Many students dive into testing examples without first recalling that the LCM of any two positive integers must be at least as large as each of the original numbers. This key property immediately eliminates certain possibilities.

3. Overlooking the "positive integers" constraint
Students may not pay attention to the fact that x and y are specifically stated to be positive integers, which means both \(\mathrm{x > 0}\) and \(\mathrm{y > 0}\). This constraint is crucial for eliminating options that could result in negative or zero values.

Errors while executing the approach

1. Testing examples unsystematically
When testing each option, students often pick random examples without considering edge cases or systematic scenarios (like when \(\mathrm{x > y}\), \(\mathrm{x < y}\), or \(\mathrm{x = y}\)). This can lead to missing counterexamples or incorrectly concluding that an option works.

2. Arithmetic errors when calculating LCM
Students may make computational mistakes when finding the LCM of their chosen examples, especially for numbers that aren't relatively prime. For instance, incorrectly calculating \(\mathrm{LCM(4,6)}\) as 24 instead of 12.

3. Stopping after finding one working example
For options (A), (B), (C), and (E), students might find one example where the option works and conclude it's possible, without thoroughly checking if it can work in general. They need to verify their reasoning applies broadly.

Errors while selecting the answer

1. Picking an option that CAN be an LCM instead of CANNOT
After testing examples, students may accidentally select an option they proved works (like \(\mathrm{xy}\) or \(\mathrm{x+y}\)) instead of the one they proved impossible (\(\mathrm{x-y}\)). This happens when they lose track of the "CANNOT be" requirement in the question.

2. Second-guessing the logical proof
Even after correctly reasoning that \(\mathrm{x-y}\) violates the LCM size constraint, students may doubt their logic and pick a "safer" option because the mathematical reasoning feels too abstract compared to concrete examples.