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The integers \(\mathrm{6}\), \(\mathrm{9}\), \(\mathrm{10}\), and \(\mathrm{25}\) are each factors of the positive integer \(\mathrm{k}\). Which of the following integers must also be a factor of \(\mathrm{k}\)?
Let's start by understanding what it means for a number to be a factor of k. If 6 is a factor of k, it means k can be divided by 6 with no remainder left over. Think of it like this: if you have k cookies and you want to divide them equally into groups of 6, you'll have exactly zero cookies left over.
Since 6, 9, 10, and 25 are ALL factors of k, this means k must be divisible by each of these numbers individually. In other words, k is some number that when divided by 6 gives a whole number, when divided by 9 gives a whole number, when divided by 10 gives a whole number, and when divided by 25 gives a whole number.
We need to figure out which of the three options (54, 60, or 90) MUST also be factors of k. The key word here is "must" - we're looking for numbers that will ALWAYS be factors of k, no matter what specific value k has (as long as it satisfies our given conditions).
Process Skill: TRANSLATE
Here's the key insight: if k must be divisible by 6, 9, 10, and 25, then k must be divisible by the smallest number that contains all of these as factors. This smallest number is called the least common multiple (LCM).
Think of it this way: imagine you're looking for the smallest number of cookies that can be divided equally into groups of 6, or groups of 9, or groups of 10, or groups of 25. That's exactly what the LCM gives us.
To find LCM(6, 9, 10, 25), let's break each number into its prime factors:
• \(6 = 2 \times 3\)
• \(9 = 3^2\)
• \(10 = 2 \times 5\)
• \(25 = 5^2\)
For the LCM, we take the highest power of each prime that appears:
• Highest power of 2: \(2^1\) (from 6 or 10)
• Highest power of 3: \(3^2\) (from 9)
• Highest power of 5: \(5^2\) (from 25)
So LCM = \(2^1 \times 3^2 \times 5^2 = 2 \times 9 \times 25 = 450\)
This means k must be at least 450, or some multiple of 450 (like 900, 1350, etc.).
Now here's the crucial question: if k must be a multiple of 450, which of our options (54, 60, 90) will ALWAYS divide k?
The answer is simple: an option will always divide k if and only if that option divides 450. Why? Because if k is any multiple of 450 (like k = 450, k = 900, k = 1350, etc.), then anything that divides 450 will automatically divide all these multiples too.
Let's check each option:
Option I: Does 54 divide 450?
\(450 \div 54 = 8.33...\)
Since this isn't a whole number, 54 does NOT always divide k.
Option II: Does 60 divide 450?
\(450 \div 60 = 7.5\)
Since this isn't a whole number, 60 does NOT always divide k.
Option III: Does 90 divide 450?
\(450 \div 90 = 5\)
Since this IS a whole number, 90 WILL always divide k.
Process Skill: INFER
Let's double-check our logic with a concrete example. We found that k could be 450. Let's verify:
• \(450 \div 6 = 75\) ✓
• \(450 \div 9 = 50\) ✓
• \(450 \div 10 = 45\) ✓
• \(450 \div 25 = 18\) ✓
Now checking our Roman numerals with k = 450:
• \(450 \div 54 = 8.33...\) ✗ (not a factor)
• \(450 \div 60 = 7.5\) ✗ (not a factor)
• \(450 \div 90 = 5\) ✓ (is a factor)
Since we found that 54 and 60 are NOT factors of at least one possible value of k (namely 450), they cannot be guaranteed factors. Only 90 must always be a factor of k.
Only Roman numeral III (90) must be a factor of k. Therefore, the answer is A: III only.
1. Misunderstanding what "must be a factor" means: Students often think they need to check if the given options are factors of the individual numbers 6, 9, 10, and 25, rather than understanding that they need to find what MUST be a factor of ANY number k that has all four given numbers as factors. This leads them down the wrong path entirely.
2. Confusing "factor" with "multiple": Some students get confused about the direction of the relationship. They might think that since 6 is a factor of k, then k must be a factor of 6, which is backwards. Remember: if 6 is a factor of k, then k is a multiple of 6, not the other way around.
3. Missing the LCM connection: Students may not realize that they need to find the LCM of the given numbers. Instead, they might try to multiply all the numbers together \(6 \times 9 \times 10 \times 25\) or use some other approach that doesn't account for overlapping prime factors.
1. Prime factorization errors: When breaking down numbers into prime factors, students commonly make mistakes like writing \(9 = 3 \times 3\) instead of \(9 = 3^2\), or forgetting to identify the highest power of each prime when calculating the LCM. For example, they might use \(3^1\) instead of \(3^2\) when both \(6 = 2 \times 3^1\) and \(9 = 3^2\) are present.
2. Arithmetic mistakes in LCM calculation: Even with correct prime factorizations, students often make computational errors when calculating \(2^1 \times 3^2 \times 5^2 = 2 \times 9 \times 25\). They might calculate this as 350 or 500 instead of the correct 450.
3. Division errors when testing factors: When checking if 54, 60, or 90 divide 450, students may make basic division mistakes. For instance, they might incorrectly calculate \(450 \div 90 = 4\) instead of 5, or round \(450 \div 54 = 8.33...\) to 8 and conclude it divides evenly.
1. Misreading Roman numeral combinations: Students often correctly identify that only option III (90) works, but then mistakenly select answer choice C (I and III only) or D (II and III only) instead of A (III only). They may rush through reading the answer choices and not carefully match their findings.
2. Including non-guaranteed factors: Even after finding that 54 and 60 are not factors of 450, some students think "well, they could be factors of some other value of k" and incorrectly include them in their answer. They miss the crucial point that the question asks what MUST be a factor, meaning it has to work for ALL possible values of k.