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When \(24\) is divided by the positive integer \(\mathrm{n}\), the remainder is \(4\). Which of the following statements about \(\mathrm{n}\) must be true?
Let's start by understanding what "\(24\) divided by \(\mathrm{n}\) with remainder \(4\)" actually means in everyday terms.
When we divide \(24\) by some number \(\mathrm{n}\) and get remainder \(4\), we're essentially saying: "If I take \(24\) objects and group them into equal piles of size \(\mathrm{n}\), I'll have some complete piles plus \(4\) objects left over."
For example, if \(\mathrm{n}\) were \(5\): \(24 \div 5 = 4\) remainder \(4\), meaning \(4\) complete groups of \(5\), plus \(4\) left over.
This gives us our mathematical relationship: \(24 = \mathrm{n} \times (\text{some whole number}) + 4\)
Two critical insights emerge:
Process Skill: TRANSLATE - Converting the remainder relationship into actionable mathematical constraints
Now let's use our understanding to find what values \(\mathrm{n}\) can actually take.
Since \(24 = \mathrm{n} \times \mathrm{q} + 4\) (where \(\mathrm{q}\) is the quotient), we can rearrange this:
\(24 - 4 = \mathrm{n} \times \mathrm{q}\)
\(20 = \mathrm{n} \times \mathrm{q}\)
This tells us that \(\mathrm{n}\) must be a factor of \(20\). In simple terms: \(\mathrm{n}\) must divide \(20\) evenly.
Why does this make sense? Because after we remove the \(4\) "leftover" objects from our \(24\) total objects, the remaining \(20\) objects must form complete groups of size \(\mathrm{n}\) with nothing left over.
Let's find all positive integers that divide \(20\) evenly and are greater than \(4\).
The factors of \(20\) are: \(1, 2, 4, 5, 10, 20\)
But remember, \(\mathrm{n}\) must be greater than \(4\) (since the remainder is \(4\)).
So our possible values are: \(\mathrm{n} = 5, 10, \text{ or } 20\)
Let's verify these work:
Process Skill: APPLY CONSTRAINTS - Using the remainder condition to limit possible values
Now we check each statement against our possible values: \(\mathrm{n} \in \{5, 10, 20\}\)
Statement I: n is even
Since \(\mathrm{n} = 5\) is odd, Statement I is not always true.
Statement II: n is a multiple of 5
All possible values of \(\mathrm{n}\) are multiples of \(5\), so Statement II is always true.
Statement III: n is a factor of 20
All possible values of \(\mathrm{n}\) are factors of \(20\), so Statement III is always true.
Process Skill: CONSIDER ALL CASES - Testing each statement against every possible value to ensure completeness
Statements II and III must be true, while Statement I is not always true.
The answer is D) II and III only.
This makes intuitive sense: since \(\mathrm{n}\) must divide the "leftover" amount of \(20\) after removing the remainder, \(\mathrm{n}\) must be a factor of \(20\) (Statement III). Since \(20 = 4 \times 5\), all factors of \(20\) that are greater than \(4\) must contain the factor \(5\), making them multiples of \(5\) (Statement II).
1. Misunderstanding the remainder constraint
Students often forget that when the remainder is \(4\), the divisor \(\mathrm{n}\) must be greater than \(4\). They might include values like \(\mathrm{n} = 2\) or \(\mathrm{n} = 4\) in their list of possibilities, not realizing that mathematically, the remainder must always be smaller than the divisor.
2. Incorrectly setting up the division relationship
Some students struggle to translate "\(24\) divided by \(\mathrm{n}\) gives remainder \(4\)" into the proper mathematical form \(24 = \mathrm{n} \times \mathrm{q} + 4\). They might try to work directly with fractions or decimals instead of using the division algorithm, missing the key insight that \(20\) must be divisible by \(\mathrm{n}\).
3. Confusing "factor of" vs "multiple of"
Students frequently mix up these concepts when setting up the problem. They might think \(\mathrm{n}\) must be a multiple of \(20\) instead of a factor of \(20\), leading them down the wrong path entirely.
1. Incomplete factor identification
When finding factors of \(20\), students sometimes miss factors (forgetting \(1, 2, 4, 5, 10, 20\)) or fail to systematically check all possibilities. They might only consider obvious factors like \(5\) and \(10\) but miss \(20\) as a valid option.
2. Arithmetic errors in verification
When checking if their values work (like verifying \(24 \div 5 = 4\) remainder \(4\)), students often make basic division errors or incorrectly calculate remainders, leading them to reject valid solutions or accept invalid ones.
3. Forgetting to apply the n > 4 constraint
Even after correctly identifying that \(\mathrm{n}\) must be a factor of \(20\), students sometimes forget to eliminate factors that are \(4\) or smaller, incorrectly including \(\mathrm{n} = 1, 2, \text{ or } 4\) in their final list of possibilities.
1. Testing statements against individual cases instead of all cases
Students might check if Statement I (\(\mathrm{n}\) is even) is true for \(\mathrm{n} = 10\) and conclude it's always true, forgetting to test it against \(\mathrm{n} = 5\). They need to verify each statement holds for ALL possible values of \(\mathrm{n}\), not just some.
2. Misreading "must be true" vs "could be true"
The question asks which statements MUST be true, but students sometimes select statements that are only true for some values of \(\mathrm{n}\). They might choose Statement I because it's true for \(\mathrm{n} = 10\) and \(\mathrm{n} = 20\), missing that it fails for \(\mathrm{n} = 5\).
3. Incorrect combination selection
Even after correctly determining that only Statements II and III are always true, students might accidentally select the wrong answer choice due to rushing or misreading the combinations (like choosing C instead of D).