If d is the greatest common divisor of x and y, where 1 and (1 , then (d is greatest...

1. Translate the problem requirements: We know \(\mathrm{d}\) is the greatest common divisor of \(\mathrm{x}\) and \(\mathrm{y}\), meaning \(\mathrm{d}\) is the largest number that divides both \(\mathrm{x}\) and \(\mathrm{y}\) evenly. We need to determine which of the three given pairs will also have \(\mathrm{d}\) as their greatest common divisor.
2. Apply GCD properties to concrete examples: Use specific numbers where we can clearly see the relationships (like \(\mathrm{x=12, y=18, d=6}\)) to test each option and understand the underlying patterns.
3. Analyze each option systematically: For each pair (I, II, III), determine what their actual GCD would be and compare it to \(\mathrm{d}\).
4. Verify the pattern holds generally: Confirm that our findings from the concrete example reflect the general mathematical relationship for any valid \(\mathrm{x}\), \(\mathrm{y}\), and \(\mathrm{d}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know in plain English. We have two numbers, \(\mathrm{x}\) and \(\mathrm{y}\), and their greatest common divisor is \(\mathrm{d}\). This means \(\mathrm{d}\) is the biggest number that divides both \(\mathrm{x}\) and \(\mathrm{y}\) evenly, with no remainder left over.

We're also told that \(\mathrm{d}\) is bigger than 1, and smaller than both \(\mathrm{x}\) and \(\mathrm{y}\). This means \(\mathrm{d}\) isn't just the trivial divisor 1, and it's a proper divisor of both numbers.

Now we need to figure out: for which of these three pairs will \(\mathrm{d}\) also be their greatest common divisor?

• I. \(\mathrm{x}\) and 1
• II. \(\mathrm{y}\) and \(\mathrm{xy}\)
• III. \(\mathrm{y}\) and \(\mathrm{x-y}\)

Process Skill: TRANSLATE - Converting the formal GCD language into everyday understanding

2. Apply GCD properties to concrete examples

Let's use specific numbers to see what's happening. Say \(\mathrm{x = 12}\) and \(\mathrm{y = 18}\). What's their greatest common divisor?

The divisors of 12 are: 1, 2, 3, 4, 6, 12

The divisors of 18 are: 1, 2, 3, 6, 9, 18

The common divisors are: 1, 2, 3, 6

So \(\mathrm{d = 6}\) (the greatest common divisor)

This fits our constraints: \(\mathrm{1 < 6 < 12}\) and \(\mathrm{1 < 6 < 18}\). Perfect!

Now let's test each option with these concrete numbers:

Process Skill: SIMPLIFY - Using concrete numbers makes the abstract concept tangible

3. Analyze each option systematically

Option I: x and 1

We're looking at the GCD of 12 and 1. What's the biggest number that divides both 12 and 1 evenly? Well, the only divisor of 1 is 1 itself. So the GCD of 12 and 1 is just 1.

Is this equal to \(\mathrm{d = 6}\)? No! So option I doesn't work.

Option II: y and xy

We're looking at the GCD of 18 and \(\mathrm{(12 \times 18 = 216)}\). Since \(\mathrm{216 = 12 \times 18}\), we know that 18 divides 216 evenly. In fact, \(\mathrm{y}\) always divides \(\mathrm{xy}\) evenly because \(\mathrm{xy = x \times y}\). This means the GCD of \(\mathrm{y}\) and \(\mathrm{xy}\) is just \(\mathrm{y}\) itself.

So the GCD of 18 and 216 is 18. Is this equal to \(\mathrm{d = 6}\)? No! So option II doesn't work.

Option III: y and x-y

We're looking at the GCD of 18 and \(\mathrm{(12 - 18 = -6)}\). Wait, that's negative. Let's think about this more carefully. If \(\mathrm{x < y}\), then \(\mathrm{x - y}\) is negative. But let's consider the absolute value and see what happens when \(\mathrm{x > y}\) instead.

Actually, let's try different numbers where \(\mathrm{x > y}\). Let \(\mathrm{x = 18}\) and \(\mathrm{y = 12}\), so \(\mathrm{d = 6}\) still.

Now \(\mathrm{x - y = 18 - 12 = 6}\). What's the GCD of 12 and 6?

Since 6 divides 12 evenly \(\mathrm{(12 = 2 \times 6)}\), the GCD of 12 and 6 is 6.

Is this equal to \(\mathrm{d = 6}\)? Yes! This looks promising.

Process Skill: CONSIDER ALL CASES - Testing different scenarios to verify the pattern

4. Verify the pattern holds generally

Let's think about why option III works in general, using plain reasoning:

If \(\mathrm{d}\) divides both \(\mathrm{x}\) and \(\mathrm{y}\), then \(\mathrm{d}\) also divides any combination like \(\mathrm{(x - y)}\). This is because if \(\mathrm{x = d \times m}\) and \(\mathrm{y = d \times n}\) for some integers \(\mathrm{m}\) and \(\mathrm{n}\), then \(\mathrm{x - y = d \times m - d \times n = d \times (m - n)}\). So \(\mathrm{d}\) definitely divides \(\mathrm{(x - y)}\).

But we need to check if \(\mathrm{d}\) is the greatest common divisor. There's a fundamental property: \(\mathrm{GCD(a, b) = GCD(a, a-b)}\). This means the greatest common divisor doesn't change when we replace one number with their difference.

So \(\mathrm{GCD(y, x-y) = GCD(x, y) = d}\).

Let's double-check our other conclusions:

• Option I: \(\mathrm{GCD(x, 1) = 1}\) always, regardless of what \(\mathrm{d}\) is
• Option II: \(\mathrm{GCD(y, xy) = y}\), which is generally not equal to \(\mathrm{d}\)
• Option III: \(\mathrm{GCD(y, x-y) = d}\) ✓

Process Skill: INFER - Drawing the non-obvious conclusion about the GCD property

5. Final Answer

Only option III gives us \(\mathrm{d}\) as the greatest common divisor.

The answer is (C) III only.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting the constraints on d

Students often overlook or misinterpret the given constraints \(\mathrm{"1 < d < x"}\) and \(\mathrm{1 < d < y"}\). They may assume \(\mathrm{d}\) can equal 1 or equal \(\mathrm{x}\) or \(\mathrm{y}\), which significantly changes the problem. This constraint tells us that \(\mathrm{d}\) is a proper divisor (not 1 or the number itself), but students frequently miss this crucial detail when setting up their approach.

Faltering Point 2: Not recognizing the need to test each option systematically

Students may try to solve this abstractly without testing concrete examples or may only test one or two options instead of systematically checking all three. This leads to incomplete analysis and potentially wrong conclusions about which options work.

Faltering Point 3: Confusing "greatest common divisor" with "common divisor"

Students might think that if \(\mathrm{d}\) divides both numbers in a pair, then \(\mathrm{d}\) is automatically the GCD. They fail to recognize that \(\mathrm{d}\) needs to be the greatest common divisor, not just any common divisor. This misunderstanding leads to incorrect evaluation of the options.

Errors while executing the approach

Faltering Point 1: Arithmetic errors when calculating GCD or applying GCD properties

When testing concrete examples (like \(\mathrm{x=12, y=18}\)), students may make computational errors in finding divisors or calculating the GCD. For instance, they might incorrectly list the divisors of 18 as "1, 2, 3, 9, 18" (missing 6) or make errors when computing \(\mathrm{x-y}\).

Faltering Point 2: Handling negative results incorrectly

When \(\mathrm{x < y}\), the expression \(\mathrm{x-y}\) becomes negative. Students may get confused about how to handle GCD with negative numbers or may incorrectly conclude that option III doesn't work because they tried cases where \(\mathrm{x-y < 0}\) without considering the absolute value or testing cases where \(\mathrm{x > y}\).

Faltering Point 3: Incorrectly applying the GCD property for option II

For option II \(\mathrm{(y \text{ and } xy)}\), students may incorrectly conclude that since \(\mathrm{d}\) divides both \(\mathrm{y}\) and \(\mathrm{xy}\), then \(\mathrm{d}\) must be their GCD. They fail to recognize that \(\mathrm{GCD(y, xy) = y}\), not \(\mathrm{d}\), because \(\mathrm{y}\) is the largest number that divides both \(\mathrm{y}\) and \(\mathrm{xy}\).

Errors while selecting the answer

Faltering Point 1: Misreading the answer choices

Students may correctly determine that only option III works but then select the wrong answer choice. For example, they might choose "(B) II only" instead of "(C) III only" due to careless reading or mixing up the Roman numerals.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose concrete values that satisfy the given conditions

We need \(\mathrm{d}\) to be the GCD of \(\mathrm{x}\) and \(\mathrm{y}\), where \(\mathrm{1 < d < x}\) and \(\mathrm{1 < d < y}\). Let's choose:

• \(\mathrm{x = 12}\)
• \(\mathrm{y = 18}\)
• \(\mathrm{d = 6}\)

We can verify: \(\mathrm{GCD(12, 18) = 6}\), and indeed \(\mathrm{1 < 6 < 12}\) and \(\mathrm{1 < 6 < 18}\) ✓

Step 2: Test each option with our concrete numbers

Option I: x and 1

\(\mathrm{GCD(12, 1) = 1}\)

Since \(\mathrm{1 \neq 6}\), option I is false.

Option II: y and xy

\(\mathrm{xy = 12 \times 18 = 216}\)

\(\mathrm{GCD(18, 216) = 18}\)

Since \(\mathrm{18 \neq 6}\), option II is false.

Option III: y and x-y

\(\mathrm{x - y = 12 - 18 = -6}\)

\(\mathrm{GCD(18, -6) = GCD(18, 6) = 6}\)

Since \(\mathrm{6 = 6}\), option III is true! ✓

Step 3: Verify with another set of smart numbers

Let's double-check with \(\mathrm{x = 15, y = 21, d = 3}\):

• \(\mathrm{GCD(15, 21) = 3}\) ✓
• Option I: \(\mathrm{GCD(15, 1) = 1 \neq 3}\) ✗
• Option II: \(\mathrm{GCD(21, 315) = 21 \neq 3}\) ✗
• Option III: \(\mathrm{GCD(21, -6) = GCD(21, 6) = 3}\) ✓

Conclusion: Only option III consistently gives us \(\mathrm{d}\) as the GCD.

Answer: (C) III only