If k and m are integers and \(\mathrm{k} = \mathrm{m} \times (\mathrm{m}+4) \times (\mathrm{m}+5)\), k must be divisible by which...

1. Translate the problem requirements: We need to determine which numbers must always divide k, where \(\mathrm{k = m \times (m+4) \times (m+5)}\) for any integer m. The Roman numerals ask whether k is always divisible by 3, 6, and 8.
2. Analyze the structure of consecutive-type integers: Recognize that m, m+4, and m+5 are consecutive integers when we consider m and m+5, giving us insights about divisibility patterns.
3. Test divisibility by 3 using modular arithmetic: Check all possible remainders when m is divided by 3 to see if the product is always divisible by 3.
4. Test divisibility by 6 and 8: Since \(\mathrm{6 = 2 \times 3}\), check if we always have both factors of 2 and 3. For \(\mathrm{8 = 2^3}\), determine if we always have three factors of 2 in the product.
5. Evaluate which statements are always true: Combine the results to determine which of I, II, and III must always hold.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to find. We have \(\mathrm{k = m \times (m+4) \times (m+5)}\) where m is any integer. The question asks which of the following numbers MUST always divide k, no matter what integer value m takes:

• I. 3
• II. 6
• III. 8

The key word here is "must" - we need to prove that k is ALWAYS divisible by these numbers, regardless of what integer m we choose.

Process Skill: TRANSLATE

2. Analyze the structure of consecutive-type integers

Let's look at the three numbers we're multiplying: m, (m+4), and (m+5). Notice that (m+4) and (m+5) are consecutive integers - they're right next to each other. Also, between m and (m+5), we have exactly 5 consecutive integers: m, m+1, m+2, m+3, m+4, m+5.

This structure gives us powerful insights about divisibility. When we have consecutive integers, we can guarantee certain divisibility patterns because consecutive numbers have predictable remainder patterns when divided by small numbers like 2, 3, etc.

3. Test divisibility by 3 using modular arithmetic

To check if k is always divisible by 3, let's consider what happens when we divide m by 3. Any integer m can have remainder 0, 1, or 2 when divided by 3.

Case 1: If \(\mathrm{m \equiv 0 \pmod{3}}\), then m is divisible by 3, so k is divisible by 3.

Case 2: If \(\mathrm{m \equiv 1 \pmod{3}}\), then:

• \(\mathrm{m+4 \equiv 1+4 \equiv 5 \equiv 2 \pmod{3}}\)
• \(\mathrm{m+5 \equiv 1+5 \equiv 6 \equiv 0 \pmod{3}}\)

So (m+5) is divisible by 3, making k divisible by 3.

Case 3: If \(\mathrm{m \equiv 2 \pmod{3}}\), then:

• \(\mathrm{m+4 \equiv 2+4 \equiv 6 \equiv 0 \pmod{3}}\)

So (m+4) is divisible by 3, making k divisible by 3.

In all cases, k is divisible by 3. So statement I is always true.

Process Skill: CONSIDER ALL CASES

4. Test divisibility by 6 and 8

Testing divisibility by 6:
Since \(\mathrm{6 = 2 \times 3}\), we need k to be divisible by both 2 and 3. We already proved k is always divisible by 3.

For divisibility by 2: Among any two consecutive integers (m+4) and (m+5), one must be even and one must be odd. So their product (m+4)(m+5) is always even, making \(\mathrm{k = m \times (m+4) \times (m+5)}\) always even.

Therefore, k is always divisible by both 2 and 3, so k is always divisible by 6. Statement II is always true.

Testing divisibility by 8:
For k to be always divisible by \(\mathrm{8 = 2^3}\), we would need at least three factors of 2 in every case. Let's test with a specific example:

If \(\mathrm{m = 1}\): \(\mathrm{k = 1 \times 5 \times 6 = 30}\)
\(\mathrm{30 = 2 \times 3 \times 5}\), which has only one factor of 2, so 30 is not divisible by 8.

Since we found a case where k is not divisible by 8, statement III is not always true.

5. Evaluate which statements are always true

From our analysis:

• Statement I (divisible by 3): Always true
• Statement II (divisible by 6): Always true
• Statement III (divisible by 8): Not always true

Therefore, statements I and II are always true.

Final Answer

The answer is B. I and II.

k must always be divisible by 3 and 6, but not necessarily by 8. We verified this through systematic case analysis for divisibility by 3, logical reasoning about consecutive integers for divisibility by 6, and a counterexample for divisibility by 8.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "must be divisible"
Students often fail to recognize that "k must be divisible by" means the divisibility should hold for ALL possible integer values of m, not just some values. They might test only a few specific cases and conclude incorrectly, missing the requirement to prove universal divisibility.

2. Not recognizing the consecutive integer pattern
Students may not notice that (m+4) and (m+5) are consecutive integers, which is crucial for establishing guaranteed divisibility patterns. Without this insight, they miss the systematic approach using modular arithmetic and may resort to random testing.

3. Confusing "which of the following" with "how many"
The question asks which statements (I, II, III) are always true, but students might think they need to find just one answer or count how many are true, rather than identifying all statements that must always hold.

Errors while executing the approach

1. Incomplete case analysis for modular arithmetic
When testing divisibility by 3, students might check only one or two cases (like \(\mathrm{m \equiv 0}\) or \(\mathrm{m \equiv 1 \pmod{3}}\)) and forget to verify the third case (\(\mathrm{m \equiv 2 \pmod{3}}\)), leading to an incomplete proof.

2. Arithmetic errors in modular calculations
Students often make mistakes when computing remainders, such as calculating \(\mathrm{(m+4) \equiv 1+4 \equiv 5 \equiv 2 \pmod{3}}\) incorrectly, or forgetting that \(\mathrm{6 \equiv 0 \pmod{3}}\), which can invalidate their entire analysis.

3. Insufficient counterexample testing
For divisibility by 8, students might test only one example (or none at all) to conclude it's not always true. They should verify their counterexample thoroughly - for instance, confirming that \(\mathrm{k = 30}\) when \(\mathrm{m = 1}\) is indeed not divisible by 8.

Errors while selecting the answer

1. Selecting based on first valid statement only
After proving that statement I (divisible by 3) is always true, students might immediately select answer choice A without checking whether statements II and III are also always true, missing that the correct answer includes multiple statements.

2. Misreading answer choices
Students might confuse the Roman numerals or answer choice letters. For example, they might identify that statements I and II are correct but accidentally select answer choice D ("II only") instead of answer choice B ("I and II").

Alternate Solutions

Smart Numbers Approach

For this divisibility problem, we can strategically test specific values of m to identify patterns, then verify our findings.

Step 1: Choose strategic test values for m
Let's test \(\mathrm{m = 0, 1, 2}\) to cover different remainder cases when divided by small numbers like 2 and 3.

Step 2: Calculate k for each test value

For \(\mathrm{m = 0}\): \(\mathrm{k = 0 \times 4 \times 5 = 0}\)
For \(\mathrm{m = 1}\): \(\mathrm{k = 1 \times 5 \times 6 = 30}\)
For \(\mathrm{m = 2}\): \(\mathrm{k = 2 \times 6 \times 7 = 84}\)

Step 3: Test divisibility patterns

Testing divisibility by 3:
• \(\mathrm{k = 0}\): \(\mathrm{0 \div 3 = 0}\) ✓
• \(\mathrm{k = 30}\): \(\mathrm{30 \div 3 = 10}\) ✓
• \(\mathrm{k = 84}\): \(\mathrm{84 \div 3 = 28}\) ✓

Testing divisibility by 6:
• \(\mathrm{k = 0}\): \(\mathrm{0 \div 6 = 0}\) ✓
• \(\mathrm{k = 30}\): \(\mathrm{30 \div 6 = 5}\) ✓
• \(\mathrm{k = 84}\): \(\mathrm{84 \div 6 = 14}\) ✓

Testing divisibility by 8:
• \(\mathrm{k = 0}\): \(\mathrm{0 \div 8 = 0}\) ✓
• \(\mathrm{k = 30}\): \(\mathrm{30 \div 8 = 3.75}\) ✗
• \(\mathrm{k = 84}\): \(\mathrm{84 \div 8 = 10.5}\) ✗

Step 4: Verify the pattern with additional strategic values
Test \(\mathrm{m = 3}\): \(\mathrm{k = 3 \times 7 \times 8 = 168}\)
• \(\mathrm{168 \div 3 = 56}\) ✓
• \(\mathrm{168 \div 6 = 28}\) ✓
• \(\mathrm{168 \div 8 = 21}\) ✓

Test \(\mathrm{m = 4}\): \(\mathrm{k = 4 \times 8 \times 9 = 288}\)
• \(\mathrm{288 \div 3 = 96}\) ✓
• \(\mathrm{288 \div 6 = 48}\) ✓
• \(\mathrm{288 \div 8 = 36}\) ✓

Step 5: Analyze the pattern
From our smart number tests:
• Divisibility by 3: Always true
• Divisibility by 6: Always true
• Divisibility by 8: Sometimes true, sometimes false

Therefore, statements I and II are always true, while III is not always true.

Answer: B (I and II)