If x is a multiple of 18 and 60, then x must be divisible by which of the following? 24...

1. Translate the problem requirements: If x is a multiple of both 18 and 60, we need to find what x must always be divisible by. This means finding what factors are guaranteed to divide any common multiple of 18 and 60.
2. Find the smallest possible value of x: Determine the least common multiple (LCM) of 18 and 60, which represents the smallest positive number that both 18 and 60 divide into.
3. Test each Roman numeral option: Check whether 24, 36, and 45 each divide into the LCM we found, since any multiple of the LCM will also be divisible by the same factors.
4. Match results to answer choices: Identify which combination of Roman numerals (I, II, III) gives us factors that always divide x.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. When we say "x is a multiple of 18 and 60," we mean that x is a number that both 18 and 60 can divide into evenly. Think of it this way: if you have x objects, you can arrange them into groups of 18 with no leftovers, AND you can also arrange those same x objects into groups of 60 with no leftovers.

The question wants to know what x MUST be divisible by. This means we need to find numbers that will ALWAYS divide evenly into x, no matter which specific multiple of both 18 and 60 we choose for x.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Find the smallest possible value of x

To find what x must always be divisible by, let's find the smallest possible value of x first. This smallest value is called the least common multiple (LCM) of 18 and 60.

Let's think about this step by step:

• First, let's break down 18 and 60 into their prime factors
• \(18 = 2 \times 9 = 2 \times 3^2\)
• \(60 = 4 \times 15 = 4 \times 3 \times 5 = 2^2 \times 3 \times 5\)

The LCM takes the highest power of each prime factor that appears:

• Highest power of 2: \(2^2\) (from 60)
• Highest power of 3: \(3^2\) (from 18)
• Highest power of 5: \(5^1\) (from 60)

So \(\mathrm{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)

This means the smallest possible value of x is 180. Any other value of x that's a multiple of both 18 and 60 will be a multiple of 180 (like 360, 540, 720, etc.).

3. Test each Roman numeral option

Now we need to check if 24, 36, and 45 each divide into 180. If they divide into 180, they'll divide into any multiple of 180.

Testing I: Does 24 divide 180?
\(180 \div 24 = 7.5\)
Since this isn't a whole number, 24 does NOT always divide x. So Roman numeral I is false.

Testing II: Does 36 divide 180?
\(180 \div 36 = 5\)
This is a whole number, so 36 DOES always divide x. Roman numeral II is true.

Testing III: Does 45 divide 180?
\(180 \div 45 = 4\)
This is a whole number, so 45 DOES always divide x. Roman numeral III is true.

Process Skill: APPLY CONSTRAINTS - Testing each condition systematically to avoid missing cases

4. Match results to answer choices

From our testing:

• I (24): False - does not always divide x
• II (36): True - always divides x
• III (45): True - always divides x

So we need the answer choice that includes II and III only.

Looking at the choices:
(A) I only - No, I is false
(B) II only - No, III is also true
(C) I and II only - No, I is false and III is missing
(D) II and III only - Yes! This matches our results
(E) I, II and III - No, I is false

Final Answer

The answer is (D) II and III only. Any number that is a multiple of both 18 and 60 must be divisible by 36 and 45, but is not necessarily divisible by 24.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "multiple of both" as "multiple of their product"
Students often think that if x is a multiple of both 18 and 60, then x must be a multiple of \(18 \times 60 = 1080\). This leads them to use 1080 as their base number instead of finding the LCM of 18 and 60. This fundamental misunderstanding makes them think x must be much larger than it actually needs to be.

2. Confusing "must be divisible by" with "could be divisible by"
The question asks what x MUST be divisible by (meaning always, for every possible value of x), but students may interpret this as asking what x COULD be divisible by. This leads to including numbers that work for some values of x but not all values, missing the universal requirement.

3. Attempting to test with arbitrary multiples instead of using LCM
Students might try to find a few random numbers that are multiples of both 18 and 60 (like 360, 720) and test divisibility with those, rather than systematically finding the LCM first. This approach is inefficient and may miss the complete picture of what x must always be divisible by.

Errors while executing the approach

1. Prime factorization errors
When breaking down 18 and 60 into prime factors, students commonly make mistakes like writing \(18 = 2 \times 3 \times 3\) instead of \(18 = 2 \times 3^2\), or incorrectly factoring \(60 = 2 \times 3 \times 10\) instead of \(60 = 2^2 \times 3 \times 5\). These errors lead to an incorrect LCM calculation.

2. LCM calculation mistakes
Even with correct prime factorizations, students often take the wrong powers when computing LCM. For instance, they might take \(2^1\) instead of \(2^2\) as the highest power of 2, or forget to include all prime factors (like omitting the factor of 5), leading to an incorrect LCM.

3. Division computation errors
When testing if 24, 36, and 45 divide 180, students may make basic arithmetic mistakes. For example, they might incorrectly calculate \(180 \div 24 = 8\) (instead of 7.5), leading them to incorrectly conclude that 24 always divides x.

Errors while selecting the answer

1. Misreading Roman numeral combinations
Students who correctly identify that II and III are true but I is false may still select the wrong answer choice because they misread the combinations. For example, they might choose (C) "I and II only" thinking it says "II and III only," or select (E) thinking that including extra true statements doesn't hurt.

2. Including false statements with true ones
Even when students correctly determine that statement I (24) is false, they may still select answer choice (E) "I, II and III" because they reason that "if II and III are definitely true, then the answer with the most true statements must be right." They fail to recognize that including a false statement makes the entire answer choice incorrect.

Alternate Solutions

Smart Numbers Approach

Step 1: Select a smart number for x
Since x must be a multiple of both 18 and 60, let's choose x = 180 (the LCM of 18 and 60). This is a logical choice because:

• \(180 \div 18 = 10\) ✓
• \(180 \div 60 = 3\) ✓
• 180 is the smallest positive number that satisfies our condition

Step 2: Test each Roman numeral option with x = 180

Option I: Is 180 divisible by 24?
\(180 \div 24 = 7.5\)
Since this is not a whole number, 180 is NOT divisible by 24.

Option II: Is 180 divisible by 36?
\(180 \div 36 = 5\)
Since this is a whole number, 180 IS divisible by 36.

Option III: Is 180 divisible by 45?
\(180 \div 45 = 4\)
Since this is a whole number, 180 IS divisible by 45.

Step 3: Verify with another smart number
Let's test x = 360 (which is 2 × 180, so also a multiple of both 18 and 60):

• \(360 \div 24 = 15\) ✓ (divisible)
• \(360 \div 36 = 10\) ✓ (divisible)
• \(360 \div 45 = 8\) ✓ (divisible)

Step 4: Analyze the results
We get different results for option I with different values of x:
• With x = 180: NOT divisible by 24
• With x = 360: IS divisible by 24

This means x is not ALWAYS divisible by 24, so option I is false.

However, both test values show that x is always divisible by 36 and 45, so options II and III are true.

Answer: D (II and III only)