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If \(\mathrm{d}\) is the greatest common divisor of \(\mathrm{x}\) and \(\mathrm{y}\), where \(1 < \mathrm{d} < \mathrm{x}\) and \(1 < \mathrm{d} < \mathrm{y}\), then \(\mathrm{d}\) is greatest common divisor of which of the following?
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?
Process Skill: TRANSLATE - Converting the formal GCD language into everyday understanding
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
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.
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.
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
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:
Process Skill: INFER - Drawing the non-obvious conclusion about the GCD property
Only option III gives us \(\mathrm{d}\) as the greatest common divisor.
The answer is (C) III only.
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.
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 \(\mathrm{xy}\).
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.
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:
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 \(\mathrm{x = 15, y = 21, d = 3}\):
Conclusion: Only option III consistently gives us \(\mathrm{d}\) as the GCD.
Answer: (C) III only