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Which of the following describes the set of all possible values of the positive integer k such that, for each positive odd integer n, the value of \(\frac{\mathrm{n}}{\mathrm{k}}\) is midway between consecutive integers?
Let's start by understanding what "midway between consecutive integers" actually means in everyday terms.
Imagine you have a number line with integers marked: ...1, 2, 3, 4, 5...
If a number is "midway between consecutive integers," it sits exactly halfway between two whole numbers that are next to each other.
For example:
So when the problem says "\(\mathrm{n/k}\) is midway between consecutive integers," it means \(\mathrm{n/k}\) must equal some whole number plus exactly \(\mathrm{0.5}\).
In other words: \(\mathrm{n/k = (some\ integer) + 0.5}\)
Process Skill: TRANSLATE - Converting the geometric concept of "midway" into the algebraic condition of having decimal part 0.5
Now let's work with what we discovered. If \(\mathrm{n/k}\) is midway between integers, then:
\(\mathrm{n/k = (some\ integer) + 0.5}\)
Let's call that integer m, so:
\(\mathrm{n/k = m + 0.5}\)
To make this easier to work with, let's rewrite \(\mathrm{0.5}\) as \(\frac{1}{2}\):
\(\mathrm{n/k = m + \frac{1}{2}}\)
Multiplying both sides by k:
\(\mathrm{n = k(m + \frac{1}{2})}\)
\(\mathrm{n = km + \frac{k}{2}}\)
\(\mathrm{n = k(m + \frac{1}{2}) = \frac{k(2m + 1)}{2}}\)
Since \(\mathrm{2m + 1}\) is always odd (because \(\mathrm{2m}\) is even, and even + 1 = odd), we can say:
\(\mathrm{n = \frac{k \times (odd\ number)}{2}}\)
This means n must equal k times an odd number, all divided by 2.
Here's the key insight: the problem says this must work for EVERY positive odd integer n.
So n can be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, ... (any odd positive integer)
From our work above, we need: \(\mathrm{n = \frac{k \times (odd\ number)}{2}}\)
Rearranging: \(\mathrm{(odd\ number) = \frac{2n}{k}}\)
Since n is odd and we need \(\mathrm{\frac{2n}{k}}\) to be an odd integer for ALL odd values of n, let's think about what this means for k.
For \(\mathrm{\frac{2n}{k}}\) to be an integer, k must divide 2n evenly.
For \(\mathrm{\frac{2n}{k}}\) to be odd, k cannot have any odd factors that would cancel with factors in 2n.
Since \(\mathrm{2n = 2 \times (odd\ number)}\), and we need \(\mathrm{\frac{2n}{k}}\) to be odd, k must contain all the factors of 2 but none of the odd factors.
This means k must be a power of 2.
But wait - let's check this with examples:
So \(\mathrm{k = 2}\) works, but \(\mathrm{k = 4}\) doesn't work for all odd n. This means k must equal exactly 2.
Process Skill: CONSIDER ALL CASES - Testing the constraint that the condition must hold for ALL positive odd integers n
We determined that k must equal 2. Let's check this against the answer choices:
The answer is D. All even prime numbers.
This works because:
1. Misinterpreting "midway between consecutive integers"
Students often confuse this with being "between any two integers" rather than specifically halfway between adjacent integers. They might think \(\mathrm{n/k}\) just needs to be a non-integer, missing that it must have a decimal part of exactly \(\mathrm{0.5}\).
2. Missing the critical constraint "for each positive odd integer n"
Many students focus on finding values of k that work for some odd integers (like \(\mathrm{n=1}\) or \(\mathrm{n=3}\)) but overlook that k must work for ALL positive odd integers. This leads them to consider values like \(\mathrm{k=4}\) or \(\mathrm{k=6}\) as valid when these only work for specific values of n.
3. Incorrectly setting up the algebraic relationship
Students may struggle to translate "midway between consecutive integers" into the correct mathematical form \(\mathrm{n/k = m + 0.5}\), instead setting up equations like \(\mathrm{n/k = m + 1}\) or \(\mathrm{n/k = \frac{(m+1)}{2}}\).
1. Incomplete testing of the "for all n" constraint
When testing whether \(\mathrm{k=4}\) works, students might only check \(\mathrm{n=1}\) (giving \(\mathrm{\frac{1}{4} = 0.25}\), which doesn't work) but fail to systematically verify that their chosen k value fails for other odd integers, leading to incorrect conclusions about which k values are valid.
2. Arithmetic errors when manipulating the midway condition
When working with \(\mathrm{n/k = m + \frac{1}{2}}\) and rearranging to get \(\mathrm{n = \frac{k(2m+1)}{2}}\), students often make algebraic mistakes, particularly when dealing with the fraction \(\frac{1}{2}\) or when multiplying through by k.
1. Selecting answer choices that include k=2 but contain additional invalid values
After correctly determining that \(\mathrm{k=2}\) works, students might choose answer C ("All positive even integers") because it includes 2, without recognizing that this choice also includes \(\mathrm{k=4, k=6}\), etc., which don't satisfy the condition for all odd n. They focus on the fact that their answer is included rather than checking if ALL values in the choice work.