What is the minimum number of marbles needed so that the marbles can be evenly divided among a group of...

1. Translate the problem requirements: We need the smallest number of marbles that can be evenly divided (no remainders) among groups of exactly 18, 24, 30, or 36 people. This means we need a number that is divisible by all four of these group sizes.
2. Identify the mathematical relationship: Since the marbles must divide evenly among any of these group sizes, we need to find the least common multiple (LCM) of 18, 24, 30, and 36.
3. Find prime factorizations to build the LCM: Break down each number into its prime factors to systematically determine what factors our answer must contain.
4. Construct the LCM using highest powers: Combine all necessary prime factors using their highest powers that appear across all four numbers to ensure divisibility by each original number.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. We have marbles that need to be divided evenly among groups of people. The key word here is "evenly" - this means every person in the group gets exactly the same number of marbles, with none left over.

Think of it like this: if you have a bag of marbles and 18 people, you want to give each person the same number of marbles and have an empty bag at the end. The same thing needs to work if you have 24 people, 30 people, or 36 people.

In mathematical terms, we need a number of marbles that leaves no remainder when divided by 18, 24, 30, or 36. This means our answer must be divisible by all four of these numbers.

Process Skill: TRANSLATE - Converting "evenly divided" to "divisible by all group sizes"

2. Identify the mathematical relationship

Now that we understand what we need, let's think about what kind of number works for multiple group sizes.

Imagine you have 72 marbles. Can these be evenly divided among 18 people? Yes - each person gets 4 marbles (\(72 \div 18 = 4\)). What about 24 people? Each gets 3 marbles (\(72 \div 24 = 3\)). So 72 works for both 18 and 24.

But we need a number that works for ALL four group sizes: 18, 24, 30, and 36. We're looking for the smallest such number because the problem asks for the "minimum."

This is exactly what mathematicians call the "least common multiple" or LCM. The LCM of several numbers is the smallest positive number that is divisible by all of them.

Process Skill: INFER - Recognizing that "minimum number divisible by all" means LCM

3. Find prime factorizations to build the LCM

To find the LCM systematically, we break each number down into its basic building blocks - prime factors. Think of this like taking apart each number to see what "ingredients" it's made of.

Let's break down each group size:

• 18 = 2 × 9 = 2 × 3 × 3 = \(2^1 \times 3^2\)
• 24 = 8 × 3 = 2 × 2 × 2 × 3 = \(2^3 \times 3^1\)
• 30 = 2 × 15 = 2 × 3 × 5 = \(2^1 \times 3^1 \times 5^1\)
• 36 = 4 × 9 = 2 × 2 × 3 × 3 = \(2^2 \times 3^2\)

Now we can see the "ingredients" clearly:

• 18 needs: \(2^1\) and \(3^2\)
• 24 needs: \(2^3\) and \(3^1\)
• 30 needs: \(2^1\), \(3^1\), and \(5^1\)
• 36 needs: \(2^2\) and \(3^2\)

4. Construct the LCM using highest powers

To build a number that's divisible by all four group sizes, we need to include enough of each prime factor to satisfy the "hungriest" number for that factor.

Looking at our prime factorizations:

• For factor 2: We need at least \(2^3\) (since 24 requires \(2^3\), which is more than the \(2^1\) needed by 18 and 30, or the \(2^2\) needed by 36)
• For factor 3: We need at least \(3^2\) (since both 18 and 36 require \(3^2\), which is more than the \(3^1\) needed by 24 and 30)
• For factor 5: We need at least \(5^1\) (since 30 requires \(5^1\), and the others don't need factor 5 at all)

Therefore: \(\mathrm{LCM} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360\)

Let's verify this works:

• \(360 \div 18 = 20\) ✓
• \(360 \div 24 = 15\) ✓
• \(360 \div 30 = 12\) ✓
• \(360 \div 36 = 10\) ✓

All divisions result in whole numbers with no remainders!

Final Answer

The minimum number of marbles needed is 360.

Looking at our answer choices, this corresponds to choice E: 360.

We can quickly verify that the other choices don't work. For example, choice D (180) fails because \(180 \div 24 = 7.5\), which means marbles can't be evenly divided among 24 people.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "evenly divided" as finding a common factor instead of LCM
Students often confuse this with finding what number can divide INTO all group sizes, rather than finding a number that IS divisible BY all group sizes. They might look for the greatest common divisor (GCD) instead of the least common multiple (LCM).

2. Missing the "minimum" constraint
Students may understand they need a number divisible by 18, 24, 30, and 36, but forget that the question asks for the MINIMUM such number. They might find any common multiple (like 720 or 1080) rather than the least common multiple.

3. Attempting to solve by trial and error with answer choices
Rather than recognizing this as an LCM problem, students may try to test each answer choice by dividing it by 18, 24, 30, and 36. While this can work, it's time-consuming and doesn't build conceptual understanding of why 360 is the answer.

Errors while executing the approach

1. Incorrect prime factorization
Students commonly make errors when breaking down numbers into prime factors. For example, writing \(24 = 2^2 \times 6\) instead of \(24 = 2^3 \times 3^1\), or factoring \(36 = 6^2\) instead of \(36 = 2^2 \times 3^2\). These errors lead to an incorrect LCM calculation.

2. Using wrong powers when constructing the LCM
Even with correct prime factorizations, students may use the smallest power of each prime instead of the highest power. For instance, taking \(2^1\) (the minimum) instead of \(2^3\) (the maximum needed), which would result in a number that doesn't work for all group sizes.

3. Arithmetic errors in final calculation
Students may correctly identify that \(\mathrm{LCM} = 2^3 \times 3^2 \times 5^1\) but make computational mistakes: calculating \(8 \times 9 = 81\) instead of 72, or \(72 \times 5 = 320\) instead of 360. These simple arithmetic errors lead to selecting the wrong answer choice.

Errors while selecting the answer

1. Failing to verify the final answer
Students may calculate LCM = 360 but forget to check that \(360 \div 18\), \(360 \div 24\), \(360 \div 30\), and \(360 \div 36\) all yield whole numbers. Without this verification, they might doubt their work and change to an incorrect answer.

2. Selecting a multiple of the correct answer
If students make an error and get a smaller number (like 180), they might recognize it doesn't work for all cases but then incorrectly choose a larger option like 720, thinking "bigger is safer." They don't realize that the question specifically asks for the minimum.