For positive integers M and N, if 24 is a divisor of M and 36 is a divisor of N,...

1. Translate the problem requirements: We need to find the largest number that MUST divide every possible sum \(\mathrm{M+N}\), where \(\mathrm{M}\) is divisible by 24 and \(\mathrm{N}\) is divisible by 36. The key word is "must" - meaning it works for ALL valid \(\mathrm{M}\) and \(\mathrm{N}\) values.
2. Express M and N in terms of their divisors: Since 24 divides \(\mathrm{M}\), we can write \(\mathrm{M = 24k}\) for some integer \(\mathrm{k}\). Since 36 divides \(\mathrm{N}\), we can write \(\mathrm{N = 36j}\) for some integer \(\mathrm{j}\).
3. Find the structure of M + N: Substitute to get \(\mathrm{M + N = 24k + 36j}\), then factor out the greatest common factor to see what always divides this sum.
4. Verify with the constraint of finding the maximum: Test if larger values from answer choices must always divide \(\mathrm{M+N}\) by checking counterexamples with small values of \(\mathrm{k}\) and \(\mathrm{j}\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're really being asked to find. We have two types of numbers:

• \(\mathrm{M}\) is any positive integer that 24 divides evenly (like 24, 48, 72, 96, etc.)
• \(\mathrm{N}\) is any positive integer that 36 divides evenly (like 36, 72, 108, 144, etc.)

The question asks: what's the greatest integer that MUST be a divisor of \(\mathrm{M + N}\)?

The key word here is "MUST" - this means we need to find the largest number that will ALWAYS divide any sum \(\mathrm{M + N}\), no matter which specific values of \(\mathrm{M}\) and \(\mathrm{N}\) we choose (as long as they meet our conditions).

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

2. Express M and N in terms of their divisors

Since we know 24 divides \(\mathrm{M}\) evenly, this means \(\mathrm{M}\) is some whole number multiplied by 24. Let's call that whole number \(\mathrm{k}\), so:

\(\mathrm{M = 24k}\) (where \(\mathrm{k}\) is any positive integer)

Similarly, since 36 divides \(\mathrm{N}\) evenly, \(\mathrm{N}\) is some whole number multiplied by 36. Let's call that whole number \(\mathrm{j}\), so:

\(\mathrm{N = 36j}\) (where \(\mathrm{j}\) is any positive integer)

For example:

• If \(\mathrm{k = 1}\), then \(\mathrm{M = 24}\)
• If \(\mathrm{k = 2}\), then \(\mathrm{M = 48}\)
• If \(\mathrm{j = 1}\), then \(\mathrm{N = 36}\)
• If \(\mathrm{j = 3}\), then \(\mathrm{N = 108}\)

3. Find the structure of M + N

Now let's see what happens when we add \(\mathrm{M}\) and \(\mathrm{N}\):

\(\mathrm{M + N = 24k + 36j}\)

To find what must always divide this sum, let's factor out the greatest common factor of 24 and 36.

First, let's find what 24 and 36 have in common:

• \(\mathrm{24 = 2^3 \times 3^1 = 8 \times 3}\)
• \(\mathrm{36 = 2^2 \times 3^2 = 4 \times 9}\)

The greatest common factor of 24 and 36 is \(\mathrm{2^2 \times 3^1 = 4 \times 3 = 12}\).

So we can factor:

\(\mathrm{M + N = 24k + 36j = 12(2k) + 12(3j) = 12(2k + 3j)}\)

This shows us that 12 will ALWAYS divide \(\mathrm{M + N}\), regardless of what positive integers \(\mathrm{k}\) and \(\mathrm{j}\) are.

Process Skill: MANIPULATE - Factoring to reveal the underlying structure

4. Verify with the constraint of finding the maximum

We found that 12 always divides \(\mathrm{M + N}\), but let's check if any larger number from our answer choices must always divide \(\mathrm{M + N}\).

Let's test with the smallest possible values: \(\mathrm{k = 1}\) and \(\mathrm{j = 1}\)

\(\mathrm{M + N = 24(1) + 36(1) = 24 + 36 = 60}\)

Now let's try \(\mathrm{k = 1}\) and \(\mathrm{j = 2}\):

\(\mathrm{M + N = 24(1) + 36(2) = 24 + 72 = 96}\)

Let's check what divides both 60 and 96:

• \(\mathrm{60 = 12 \times 5}\)
• \(\mathrm{96 = 12 \times 8}\)

Since 5 and 8 share no common factors other than 1, the greatest number that divides both 60 and 96 is 12.

This confirms that while 12 always divides \(\mathrm{M + N}\), larger numbers like 24, 36, 60, or 72 do not always divide \(\mathrm{M + N}\).

Process Skill: CONSIDER ALL CASES - Testing with concrete examples to verify our general result

Final Answer

The greatest integer that must be a divisor of \(\mathrm{M + N}\) is 12.

Answer: A

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "must be a divisor" as "could be a divisor"

Students often overlook the critical word "MUST" in the question. They might look for any common divisor or the largest possible divisor in specific cases, rather than finding the greatest number that will ALWAYS divide \(\mathrm{M + N}\) for ALL possible values of \(\mathrm{M}\) and \(\mathrm{N}\). This leads them to choose larger answer choices like 60 or 72 based on specific examples.

2. Confusing the constraint with the question being asked

Some students might misread the problem and think they need to find what divides \(\mathrm{M}\) or \(\mathrm{N}\) individually, rather than what divides \(\mathrm{M + N}\). This could lead them to choose 24 (thinking about \(\mathrm{M}\)) or 36 (thinking about \(\mathrm{N}\)) instead of analyzing the sum \(\mathrm{M + N}\).

3. Not recognizing the need to use algebraic representation

Students might try to solve this problem by testing only specific numerical examples without setting up the general algebraic forms \(\mathrm{M = 24k}\) and \(\mathrm{N = 36j}\). This approach makes it nearly impossible to find what MUST always be true and often leads to incorrect conclusions based on limited examples.

Errors while executing the approach

1. Incorrectly finding the GCD of 24 and 36

When factoring \(\mathrm{M + N = 24k + 36j}\), students might make arithmetic errors in finding GCD(24, 36). Common mistakes include getting 6 instead of 12, or incorrectly computing the prime factorizations (\(\mathrm{24 = 2^3 \times 3}\) and \(\mathrm{36 = 2^2 \times 3^2}\)).

2. Incomplete factoring of the expression

Even if students correctly identify that GCD(24, 36) = 12, they might fail to properly factor out 12 from the expression \(\mathrm{24k + 36j}\). They might write something like \(\mathrm{12(2k + 36j)}\) instead of \(\mathrm{12(2k + 3j)}\), missing the complete simplification.

Errors while selecting the answer

1. Choosing based on specific examples rather than general proof

Students might test a few specific values (like \(\mathrm{M = 24, N = 36}\) giving \(\mathrm{M + N = 60}\)) and then select 60 from the answer choices because it works for their examples, without verifying that 60 must ALWAYS divide \(\mathrm{M + N}\). They fail to test whether their chosen answer works for ALL possible values.

2. Selecting the largest common divisor found in examples instead of the guaranteed divisor

When testing examples, if students find that both 60 and 96 are possible values of \(\mathrm{M + N}\), they might notice that both are divisible by 12 but then incorrectly choose a larger answer choice like 24 because 24 divides 60, without checking that 24 does not divide 96.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose smart values for M and N

Since we need \(\mathrm{M}\) to be divisible by 24 and \(\mathrm{N}\) to be divisible by 36, let's choose the smallest possible values to start our analysis:

• Let \(\mathrm{M = 24}\) (the smallest positive integer divisible by 24)
• Let \(\mathrm{N = 36}\) (the smallest positive integer divisible by 36)

Step 2: Calculate M + N for our smart numbers

\(\mathrm{M + N = 24 + 36 = 60}\)

This tells us that 60 must be divisible by our answer. Looking at the choices, this eliminates option E (72).

Step 3: Test with the next logical smart numbers

To find what MUST always divide \(\mathrm{M + N}\), we need to test other valid combinations:

• Let \(\mathrm{M = 48}\) (next multiple of 24) and \(\mathrm{N = 36}\)
• \(\mathrm{M + N = 48 + 36 = 84}\)

Now we need to find what divides both 60 and 84. Let's check our remaining answer choices:

• Does 36 divide 84? \(\mathrm{84 \div 36 = 2.33...}\) No
• Does 24 divide 84? \(\mathrm{84 \div 24 = 3.5}\) No
• Does 12 divide both 60 and 84? \(\mathrm{60 \div 12 = 5}\) ✓, \(\mathrm{84 \div 12 = 7}\) ✓

Step 4: Verify with one more set of smart numbers

Let \(\mathrm{M = 24}\) and \(\mathrm{N = 72}\) (another multiple of 36):

\(\mathrm{M + N = 24 + 72 = 96}\)

Does 12 divide 96? \(\mathrm{96 \div 12 = 8}\) ✓

Step 5: Confirm our pattern

From our smart number examples: 60, 84, and 96 are all divisible by 12. We can see that 12 consistently divides \(\mathrm{M + N}\), while larger values like 24 and 36 do not always divide every possible sum.

Answer: A (12)