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If \(\mathrm{x}\) and \(\mathrm{y}\) are different prime numbers, each greater than \(\mathrm{2}\), which of the following must be true?
Let's start by understanding what we're being asked to find. We have two different prime numbers, both greater than 2, and we need to determine which statements MUST be true for ANY such pair.
The key word here is "must" - this means we're looking for properties that are always true, not just sometimes true. If we can find even one counterexample where a statement fails, then that statement is not something that "must" be true.
Let's think about some examples of prime numbers greater than 2: 3, 5, 7, 11, 13, 17, 19, 23, etc. We'll use pairs like (3,5), (5,7), (7,11) to test our statements.
Process Skill: TRANSLATE - Converting the problem language "must be true" into the mathematical requirement that the property holds for ALL possible pairs
Here's the fundamental insight that will help us analyze all three statements: Every prime number greater than 2 must be an odd number.
Why is this true? Well, think about it in plain English: if a number greater than 2 is even, then it's divisible by 2, which means it has 2 as a factor. But if it has 2 as a factor, it can't be prime (since prime numbers only have exactly two factors: 1 and themselves).
So our key insight is: x and y are both odd numbers
This simple fact will be the foundation for analyzing each statement.
Now let's examine each Roman numeral statement using our key insight that both x and y are odd.
When we add two odd numbers, what do we get? Let's think through this with simple examples:
In general: odd + odd = even. This is because odd numbers can be written as \(\mathrm{2k+1}\) for some integer k, so \(\mathrm{(2j+1) + (2k+1) = 2(j+k+1)}\), which is even.
Now, is 91 even or odd? \(\mathrm{91 = 90 + 1}\), so 91 is odd.
Since \(\mathrm{x + y}\) must be even (because it's the sum of two odd numbers), and 91 is odd, we can never have \(\mathrm{x + y = 91}\). Therefore, \(\mathrm{x + y ≠ 91}\) must always be true.
When we subtract one odd number from another odd number, what do we get? Let's check:
In general: odd - odd = even. This is because \(\mathrm{(2j+1) - (2k+1) = 2(j-k)}\), which is even.
So Statement II must always be true.
For \(\mathrm{x/y}\) to be an integer, x would need to be divisible by y. But since both x and y are prime numbers and they are different, neither can be divisible by the other.
Here's why: if x were divisible by y, then y would be a factor of x. But since x is prime, its only factors are 1 and x itself. Since \(\mathrm{y ≠ x}\) (they're different) and \(\mathrm{y > 2 > 1}\), y cannot be a factor of x.
Therefore, \(\mathrm{x/y}\) can never be an integer, so Statement III must always be true.
Process Skill: CONSIDER ALL CASES - Systematically checking each statement against the fundamental property of odd numbers
Let's double-check our reasoning with specific examples using the prime pair (7, 11):
Statement I: \(\mathrm{7 + 11 = 18 ≠ 91}\) ✓
Statement II: \(\mathrm{7 - 11 = -4}\) (even) ✓ and \(\mathrm{11 - 7 = 4}\) (even) ✓
Statement III: \(\mathrm{7/11}\) is not an integer ✓ and \(\mathrm{11/7}\) is not an integer ✓
Let's try another pair (5, 13):
Statement I: \(\mathrm{5 + 13 = 18 ≠ 91}\) ✓
Statement II: \(\mathrm{5 - 13 = -8}\) (even) ✓ and \(\mathrm{13 - 5 = 8}\) (even) ✓
Statement III: \(\mathrm{5/13}\) is not an integer ✓ and \(\mathrm{13/5}\) is not an integer ✓
Our concrete examples confirm that all three statements (I, II, and III) must be true.
Since we've shown that statements I, II, and III must all be true for any two different prime numbers greater than 2, the answer is (E) I, II, and III.
This matches the given correct answer, confirming our systematic approach using the fundamental property that all primes greater than 2 are odd numbers.
Faltering Point 1: Missing the key insight that all primes greater than 2 are odd
Many students jump directly into testing specific number pairs without recognizing the fundamental property that every prime number greater than 2 must be odd. Without this insight, they may try to analyze each statement independently using trial and error, missing the elegant unified approach that makes all three statements easy to prove.
Faltering Point 2: Misunderstanding "must be true" vs "could be true"
Students often confuse questions asking what "must be true" with what "could be true." They might find one example where a statement works and conclude it's always true, or find one example where it doesn't work and conclude it's never true. The key is understanding that "must be true" means the statement holds for ALL possible pairs of different primes greater than 2.
Faltering Point 3: Attempting to test all statements with the same specific examples
Some students pick one or two specific prime pairs (like 3,5) and test all statements only with those examples. They may not realize that to prove something "must be true," they need either a general mathematical proof or systematic testing across multiple diverse examples to build confidence in the pattern.
Faltering Point 1: Arithmetic errors when working with odd number properties
When applying the rule that "odd + odd = even" or "odd - odd = even," students may make basic arithmetic mistakes or incorrectly apply these properties. For example, they might incorrectly conclude that 91 is even, or make errors when computing specific examples like \(\mathrm{7 + 11}\) or \(\mathrm{13 - 5}\).
Faltering Point 2: Incomplete analysis of Statement III regarding divisibility
For Statement III (\(\mathrm{x/y}\) is not an integer), students may not fully grasp why different prime numbers cannot divide each other. They might test a few examples without understanding the deeper reason: since both numbers are prime, each has only two factors (1 and itself), so neither can be a factor of the other when they're different.
Faltering Point 3: Forgetting to check both directions for Statement II
When analyzing "\(\mathrm{x - y}\) is an even integer," students might only consider \(\mathrm{x > y}\) scenarios and forget that \(\mathrm{y - x}\) would also be even (though negative). While this doesn't affect the correctness since \(\mathrm{|x - y|}\) is what matters, incomplete analysis can create doubt about whether the statement always holds.
Faltering Point 1: Misreading the Roman numeral combinations in answer choices
Students may correctly determine that statements I, II, and III are all true but then misread the answer choices. They might accidentally select (D) "II and III only" instead of (E) "I, II, and III" due to rushing or not carefully checking which combination corresponds to "all three statements."
Faltering Point 2: Second-guessing due to Statement I seeming "too specific"
Statement I (\(\mathrm{x + y ≠ 91}\)) might seem suspiciously specific compared to the more general-sounding statements II and III. Students may doubt their analysis and incorrectly conclude that such a specific statement couldn't always be true, leading them to select an answer choice that excludes Statement I.
Strategy: Select specific prime numbers greater than 2 and test each statement to verify they must always be true.
Step 1: Choose smart numbers
Let's select two different primes greater than 2:
\(\mathrm{x = 3, y = 5}\) (both are typical small primes > 2)
Step 2: Test Statement I - \(\mathrm{x + y ≠ 91}\)
\(\mathrm{3 + 5 = 8 ≠ 91}\) ✓
Let's try other pairs to see if we can ever get 91:
- For \(\mathrm{x + y = 91}\), we'd need two primes that sum to 91
- Since 91 is odd, we'd need one even prime and one odd prime
- The only even prime is 2, but we're told both primes > 2
- All primes > 2 are odd, so odd + odd = even ≠ 91 (which is odd)
Statement I is TRUE.
Step 3: Test Statement II - \(\mathrm{x - y}\) is an even integer
\(\mathrm{3 - 5 = -2}\) (even) ✓
\(\mathrm{5 - 3 = 2}\) (even) ✓
Let's verify with another pair: \(\mathrm{x = 7, y = 11}\)
\(\mathrm{7 - 11 = -4}\) (even) ✓
\(\mathrm{11 - 7 = 4}\) (even) ✓
Since all primes > 2 are odd: odd - odd = even always
Statement II is TRUE.
Step 4: Test Statement III - \(\mathrm{x/y}\) is not an integer
\(\mathrm{3/5 = 0.6}\) (not an integer) ✓
\(\mathrm{5/3 = 1.67...}\) (not an integer) ✓
Try \(\mathrm{x = 7, y = 11}\): \(\mathrm{7/11}\) and \(\mathrm{11/7}\) are both non-integers ✓
Since x and y are different primes, neither can divide the other evenly
Statement III is TRUE.
Step 5: Verify with additional smart numbers
Let's confirm with \(\mathrm{x = 13, y = 17}\):
I. \(\mathrm{13 + 17 = 30 ≠ 91}\) ✓
II. \(\mathrm{13 - 17 = -4}\) (even) ✓
III. \(\mathrm{13/17}\) is not an integer ✓
Conclusion: All statements I, II, and III must be true.
Answer: (E) I, II, and III