For all positive integers x and y, the expression xThetay is defined as the least multiple of y that is...

1. Translate the problem requirements: The expression \(\mathrm{x}\Theta\mathrm{y}\) means "the smallest multiple of \(\mathrm{y}\) that is greater than or equal to \(\mathrm{x}\)." We need \(20\Theta\mathrm{k} = 30\), meaning the smallest multiple of \(\mathrm{k}\) that is ≥ 20 must equal 30.
2. Establish the constraint relationship: For \(20\Theta\mathrm{k} = 30\) to be true, 30 must be a multiple of \(\mathrm{k}\), and the multiple of \(\mathrm{k}\) just before 30 must be less than 20.
3. Find all factors of 30: Since 30 must be a multiple of \(\mathrm{k}\), we need \(\mathrm{k}\) to be a factor of 30.
4. Test each factor against the constraint: For each factor \(\mathrm{k}\) of 30, check if the previous multiple of \(\mathrm{k}\) is less than 20, ensuring 30 is indeed the smallest multiple ≥ 20.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what the expression \(\mathrm{x}\Theta\mathrm{y}\) means in everyday language. The definition tells us that \(\mathrm{x}\Theta\mathrm{y}\) is "the least multiple of \(\mathrm{y}\) that is greater than or equal to \(\mathrm{x}\)."

Let's verify this with the given examples:

- \(2\Theta 3\): We need the smallest multiple of 3 that is ≥ 2. The multiples of 3 are: 3, 6, 9, 12... The smallest one that's ≥ 2 is 3. So \(2\Theta 3 = 3\) ✓

- \(3\Theta 2\): We need the smallest multiple of 2 that is ≥ 3. The multiples of 2 are: 2, 4, 6, 8... The smallest one that's ≥ 3 is 4. So \(3\Theta 2 = 4\) ✓

Now for our problem: \(20\Theta\mathrm{k} = 30\) means "the smallest multiple of \(\mathrm{k}\) that is greater than or equal to 20 must equal 30."

Process Skill: TRANSLATE - Converting the symbolic definition into clear mathematical understanding

2. Establish the constraint relationship

For \(20\Theta\mathrm{k} = 30\) to be true, we need two conditions:

1. 30 must be a multiple of \(\mathrm{k}\) (since our answer is 30)

2. The multiple of \(\mathrm{k}\) that comes just before 30 must be less than 20

Why the second condition? If there were a multiple of \(\mathrm{k}\) between 20 and 30, then that multiple (not 30) would be the smallest multiple of \(\mathrm{k}\) that's ≥ 20.

For example, if \(\mathrm{k} = 5\), then multiples of 5 are: 5, 10, 15, 20, 25, 30...
Since 20 is a multiple of 5 and 20 ≥ 20, we'd have \(20\Theta 5 = 20\), not 30.

Process Skill: INFER - Drawing the non-obvious conclusion about what must be true

3. Find all factors of 30

Since 30 must be a multiple of \(\mathrm{k}\), this means \(\mathrm{k}\) must be a factor of 30.

Let's find all factors of 30:
\(30 = 2 \times 3 \times 5\)

The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30

4. Test each factor against the constraint

Now we check each factor \(\mathrm{k}\) to see if the multiple of \(\mathrm{k}\) just before 30 is less than 20:

\(\mathrm{k} = 1\): Multiples: 1, 2, 3, ..., 29, 30
The multiple before 30 is 29. Since 29 ≥ 20, we'd have \(20\Theta 1 = 20\), not 30. ✗

\(\mathrm{k} = 2\): Multiples: 2, 4, 6, ..., 28, 30
The multiple before 30 is 28. Since 28 ≥ 20, we'd have \(20\Theta 2 = 20\), not 30. ✗

\(\mathrm{k} = 3\): Multiples: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
The multiple before 30 is 27. Since 27 ≥ 20, we'd have \(20\Theta 3 = 21\), not 30. ✗

\(\mathrm{k} = 5\): Multiples: 5, 10, 15, 20, 25, 30
The multiple before 30 is 25. Since 25 ≥ 20, we'd have \(20\Theta 5 = 20\), not 30. ✗

\(\mathrm{k} = 6\): Multiples: 6, 12, 18, 24, 30
The multiple before 30 is 24. Since 24 ≥ 20, we'd have \(20\Theta 6 = 24\), not 30. ✗

\(\mathrm{k} = 10\): Multiples: 10, 20, 30
The multiple before 30 is 20. Since 20 ≥ 20, we'd have \(20\Theta 10 = 20\), not 30. ✗

\(\mathrm{k} = 15\): Multiples: 15, 30
The multiple before 30 is 15. Since 15 < 20, the smallest multiple of 15 that's ≥ 20 is indeed 30. So \(20\Theta 15 = 30\). ✓

\(\mathrm{k} = 30\): Multiples: 30
There's no multiple before 30, so 30 is the first multiple. Since 30 ≥ 20, we have \(20\Theta 30 = 30\). ✓

Process Skill: CONSIDER ALL CASES - Systematically checking each possibility

5. Final Answer

Only \(\mathrm{k} = 15\) and \(\mathrm{k} = 30\) satisfy our requirement that \(20\Theta\mathrm{k} = 30\).

Therefore, there are exactly two different positive integers \(\mathrm{k}\) for which \(20\Theta\mathrm{k} = 30\).

The answer is B. Two.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the definition of \(\mathrm{x}\Theta\mathrm{y}\)
Students often misinterpret "the least multiple of \(\mathrm{y}\) that is greater than or equal to \(\mathrm{x}\)" as simply finding any multiple of \(\mathrm{y}\) around \(\mathrm{x}\), rather than specifically the smallest one that meets or exceeds \(\mathrm{x}\). This leads to incorrect setup of the problem constraints.

Faltering Point 2: Missing the critical constraint about intermediate multiples
Many students recognize that \(\mathrm{k}\) must divide 30, but fail to realize that there cannot be any multiple of \(\mathrm{k}\) between 20 and 30. They might stop after finding all factors of 30 without testing whether each factor actually satisfies the complete definition of the operation.

Faltering Point 3: Confusing "greater than or equal to" with "greater than"
Some students misread the definition and think \(\mathrm{x}\Theta\mathrm{y}\) requires the multiple to be strictly greater than \(\mathrm{x}\), leading them to exclude cases where the multiple exactly equals \(\mathrm{x}\) (like when testing \(\mathrm{k}=10\) where 20 is exactly a multiple of 10).

Errors while executing the approach

Faltering Point 1: Incomplete listing of factors of 30
When finding factors of 30, students may miss some factors (especially 6 and 15) or make arithmetic errors in the factorization process, leading to testing an incomplete set of possible values for \(\mathrm{k}\).

Faltering Point 2: Incorrect identification of the "previous multiple"
When testing each factor \(\mathrm{k}\), students may incorrectly identify what the multiple of \(\mathrm{k}\) just before 30 is. For example, when \(\mathrm{k}=15\), they might think the previous multiple is 0 instead of 15, or make errors in generating the sequence of multiples.

Faltering Point 3: Computational errors in evaluating \(20\Theta\mathrm{k}\)
Students may correctly understand the process but make mistakes when actually computing what \(20\Theta\mathrm{k}\) equals for each value of \(\mathrm{k}\), particularly for cases like \(\mathrm{k}=3\) where they need to find that \(20\Theta 3 = 21\), not 30.

Errors while selecting the answer

Faltering Point 1: Counting the wrong items
After correctly identifying that \(\mathrm{k}=15\) and \(\mathrm{k}=30\) work, some students might accidentally count all the factors they tested (8 factors) instead of counting only the ones that satisfy the condition (2 factors), leading to selecting a much higher answer choice.

Faltering Point 2: Including incorrect values in the final count
Students might make errors in their testing phase but still attempt to count their "successful" cases. For instance, if they incorrectly concluded that \(\mathrm{k}=10\) works in addition to \(\mathrm{k}=15\) and \(\mathrm{k}=30\), they would select "Three" instead of "Two."