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If \(\mathrm{k}\) is an integer, and \(\frac{35^2-1}{\mathrm{k}}\) is an integer, then \(\mathrm{k}\) could be each of the following, EXCEPT
Let's start by understanding what we're being asked. We have the expression \(\frac{35^2 - 1}{k}\), and we're told this equals an integer when k is an integer. This means k must divide evenly into \(35^2 - 1\) - in other words, k must be a divisor of \(35^2 - 1\).
The question asks which value k could NOT be, so we need to find which answer choice is NOT a divisor of \(35^2 - 1\).
First, let's calculate what \(35^2 - 1\) actually equals:
\(35^2 = 35 \times 35 = 1225\)
So \(35^2 - 1 = 1225 - 1 = 1224\)
Process Skill: TRANSLATE - Converting the divisibility language into a concrete mathematical task
Rather than working with the large number 1,224 directly, let's use a powerful algebraic pattern. The expression \(35^2 - 1\) fits the form \(a^2 - 1\), which is called a "difference of squares."
In everyday terms, when you have "something squared minus 1," you can always break it down as "(something minus 1) times (something plus 1)."
So: \(35^2 - 1 = (35 - 1)(35 + 1) = 34 \times 36\)
Let's verify: \(34 \times 36 = 1224\) ✓
This factorization makes our work much easier because now we can find the prime factors of two smaller numbers instead of one large number.
Now we need to break down both 34 and 36 into their prime factors:
For 34:
\(34 = 2 \times 17\)
For 36:
\(36 = 4 \times 9 = 2^2 \times 3^2\)
Therefore: \(35^2 - 1 = 34 \times 36 = (2 \times 17) \times (2^2 \times 3^2) = 2^3 \times 3^2 \times 17\)
In plain English, 1,224 is built from exactly these prime building blocks: three 2's, two 3's, and one 17.
Any number that divides 1,224 must be made up of only these prime factors, and it cannot use more of any prime than is available.
Now let's check each answer choice to see if it can be formed using only the available prime factors \(2^3 \times 3^2 \times 17\):
Choice A: k = 8
\(8 = 2^3\)
We have exactly \(2^3\) available, so 8 works ✓
Choice B: k = 9
\(9 = 3^2\)
We have exactly \(3^2\) available, so 9 works ✓
Choice C: k = 12
\(12 = 4 \times 3 = 2^2 \times 3\)
We need \(2^2\) (we have \(2^3\)) and \(3^1\) (we have \(3^2\)), so 12 works ✓
Choice D: k = 16
\(16 = 2^4\)
We would need four 2's, but we only have three 2's available \(2^3\)
Therefore, 16 does NOT work ✗
Choice E: k = 17
17 is prime and we have exactly one 17 available, so 17 works ✓
Process Skill: APPLY CONSTRAINTS - Systematically checking that each factor doesn't exceed what's available
The answer is D. 16
16 cannot be a value of k because \(16 = 2^4\), which requires four factors of 2, but \(35^2 - 1 = 2^3 \times 3^2 \times 17\) contains only three factors of 2. Therefore, 16 does not divide evenly into \(35^2 - 1\), making the expression \(\frac{35^2 - 1}{16}\) a non-integer.
1. Misunderstanding the divisibility requirement
Students often confuse the condition "\(\frac{35^2 - 1}{k}\) is an integer" and think they need to find values that make the expression equal to a specific integer, rather than understanding that k must be a divisor of \(35^2 - 1\). This leads them to attempt direct division or guess-and-check methods instead of finding the prime factorization.
2. Missing the "EXCEPT" in the question
A critical reading error occurs when students overlook that the question asks which value k could NOT be. They end up selecting a value that DOES work as a divisor instead of the one that doesn't, leading to choosing A, B, C, or E instead of D.
3. Attempting direct calculation instead of using algebraic patterns
Many students immediately calculate \(35^2 = 1225\) and then 1,224, and try to work with this large number directly instead of recognizing the difference of squares pattern \(a^2 - 1 = (a-1)(a+1)\). This makes the problem much harder and more prone to computational errors.
1. Incorrect prime factorization
When breaking down \(34 \times 36\), students commonly make errors such as:
- Factoring 36 as \(6 \times 6 = 6^2\) instead of \(2^2 \times 3^2\)
- Missing that \(34 = 2 \times 17\) and incorrectly writing it as having other prime factors
- Combining the factors incorrectly, leading to wrong final factorization like \(2^2 \times 3^2 \times 17\) instead of \(2^3 \times 3^2 \times 17\)
2. Arithmetic errors in basic calculations
Students may make simple computational mistakes such as:
- Calculating \(35^2\) incorrectly (getting something other than 1,225)
- Computing \(34 \times 36\) wrong when verifying the factorization
- Making errors when finding powers (e.g., writing \(2^3\) as 6 instead of 8)
1. Incorrectly counting available prime factors
When checking if \(16 = 2^4\) works, students might:
- Miscount how many factors of 2 are available in \(2^3 \times 3^2 \times 17\) (thinking they have 4 instead of 3)
- Confuse themselves about whether \(2^4\) requires more or fewer factors than what's available
- Make the error in reverse - correctly identifying that 16 needs \(2^4\) but incorrectly concluding that this IS available
2. Selecting the first non-working answer found during checking
If students check the answers out of order and find that one of the working answers (like A, B, C, or E) appears to not work due to their computational errors, they might select that answer instead of systematically checking all options to find that D is the actual exception.