If k is an integer greater than 6, all of the following must be divisible by 3 EXCEPT

1. Translate the problem requirements: We need to find which expression is NOT always divisible by 3 when \(\mathrm{k}\) is any integer greater than 6. "Must be divisible by 3" means the result is always a multiple of 3 regardless of which specific integer \(\mathrm{k}\) we choose.
2. Apply the consecutive integer divisibility principle: Recognize that among any set of consecutive integers, at least one will be divisible by 3, making their product divisible by 3.
3. Test each answer choice systematically: For expressions with consecutive or near-consecutive terms, apply the principle directly. For others, test with concrete values of \(\mathrm{k}\) to find counterexamples.
4. Verify the exception through concrete examples: Once we identify a suspicious expression, test it with multiple values of \(\mathrm{k}\) to confirm it's not always divisible by 3.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. The question asks which expression is NOT always divisible by 3 when \(\mathrm{k}\) is any integer greater than 6.

In everyday terms, this means: no matter which integer we pick for \(\mathrm{k}\) (as long as it's bigger than 6), four of these expressions will always give us a multiple of 3, but one expression will sometimes give us a number that's NOT a multiple of 3.

To find the exception, we need to test each expression and see if we can find even one value of \(\mathrm{k}\) that makes the expression NOT divisible by 3.

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

2. Apply the consecutive integer divisibility principle

Here's a key insight that makes this problem much easier: whenever we multiply consecutive integers (like \(\mathrm{7×8×9}\) or \(\mathrm{5×6}\)), at least one of those numbers must be divisible by 3.

Why? Because every third integer is divisible by 3. So in any group of consecutive integers, we're guaranteed to hit a multiple of 3.

For example:

• In \(\mathrm{7×8×9}\), the number 9 is divisible by 3
• In \(\mathrm{5×6×7}\), the number 6 is divisible by 3
• In \(\mathrm{10×11×12}\), the number 12 is divisible by 3

This means any expression that involves consecutive integers will automatically be divisible by 3.

3. Test each answer choice systematically

Let's examine each expression to see if it involves consecutive integers or if we can find a counterexample:

Choice B: \(\mathrm{3k^3}\)
This has a factor of 3 right in front, so it's always divisible by 3. No need to test further.

Choice C: \(\mathrm{(k+1)(k+5)(k+6)}\)
Let's check if \(\mathrm{k+5}\) and \(\mathrm{k+6}\) being consecutive helps us. These are consecutive integers, so their product \(\mathrm{(k+5)(k+6)}\) is always divisible by either 2 or 3. But we need to be more careful here.

Choice D: \(\mathrm{(k+2)(k-2)(k+3)}\)
This doesn't have consecutive integers, but let's test with \(\mathrm{k=7}\):
\(\mathrm{(7+2)(7-2)(7+3) = 9×5×10 = 450}\)
\(\mathrm{450 ÷ 3 = 150}\), so this works.

Choice E: \(\mathrm{k(k+1)(k+2)}\)
These are three consecutive integers: \(\mathrm{k}\), \(\mathrm{k+1}\), and \(\mathrm{k+2}\). By our principle, this must always be divisible by 3.

Choice A: \(\mathrm{k(k+3)(k-1)}\)
Let's rearrange this as \(\mathrm{(k-1)×k×(k+3)}\). We have \(\mathrm{k-1}\), \(\mathrm{k}\), and \(\mathrm{k+3}\). Notice that \(\mathrm{k-1}\) and \(\mathrm{k}\) are consecutive, but \(\mathrm{k+3}\) is not consecutive to either. Let's test this with a specific value.

Let \(\mathrm{k = 7}\):
\(\mathrm{7×10×6 = 420}\)
\(\mathrm{420 ÷ 3 = 140}\) ✓

Let \(\mathrm{k = 8}\):
\(\mathrm{8×11×7 = 616}\)
\(\mathrm{616 ÷ 3 = 205.33...}\) ✗

We found a counterexample! When \(\mathrm{k = 8}\), the expression gives us 616, which is not divisible by 3.

4. Verify the exception through concrete examples

Let's double-check our finding with \(\mathrm{k = 8}\):
\(\mathrm{k(k+3)(k-1) = 8×11×7}\)

Let's verify each factor's divisibility by 3:

• \(\mathrm{8 ÷ 3 = 2}\) remainder 2 (not divisible by 3)
• \(\mathrm{11 ÷ 3 = 3}\) remainder 2 (not divisible by 3)
• \(\mathrm{7 ÷ 3 = 2}\) remainder 1 (not divisible by 3)

Since none of the three factors is divisible by 3, their product cannot be divisible by 3.

\(\mathrm{8×11×7 = 616}\)
\(\mathrm{616 ÷ 3 = 205.33...}\)

This confirms that Choice A is not always divisible by 3.

Let's verify one more case with \(\mathrm{k = 11}\):
\(\mathrm{k(k+3)(k-1) = 11×14×10 = 1540}\)
\(\mathrm{1540 ÷ 3 = 513.33...}\)

Another counterexample confirmed.

Final Answer

The answer is Choice A: \(\mathrm{k(k+3)(k-1)}\)

This expression is not always divisible by 3. We demonstrated this with concrete examples (\(\mathrm{k = 8}\) and \(\mathrm{k = 11}\)) where the result is not a multiple of 3, while all other choices are always divisible by 3 for any integer \(\mathrm{k > 6}\).

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting "EXCEPT" in the question stem
Students often miss the word "EXCEPT" and look for expressions that ARE always divisible by 3, rather than the one that is NOT always divisible by 3. This leads them to select choice E (consecutive integers) thinking it's correct, when they should be looking for the exception.

Faltering Point 2: Forgetting the constraint "\(\mathrm{k > 6}\)"
Students may test values like \(\mathrm{k = 1, 2,}\) or 3 without realizing these violate the given constraint. This can lead to incorrect conclusions about divisibility patterns, especially since smaller values of \(\mathrm{k}\) might coincidentally make certain expressions divisible by 3.

Faltering Point 3: Not recognizing the consecutive integer divisibility principle
Many students don't realize that any product of consecutive integers is automatically divisible by 3. Instead, they attempt to test every single choice with multiple values, making the problem much more time-consuming and error-prone.

Errors while executing the approach

Faltering Point 1: Arithmetic errors in multiplication
When testing \(\mathrm{k = 8}\) in choice A: \(\mathrm{8 × 11 × 7}\), students may incorrectly calculate this as 626 instead of 616, or make other multiplication errors. Since \(\mathrm{626 ÷ 3 ≈ 208.67}\) while \(\mathrm{616 ÷ 3 ≈ 205.33}\), both show non-divisibility, but the wrong calculation could cause doubt.

Faltering Point 2: Incorrectly identifying consecutive integers
Students might mistakenly think that \(\mathrm{k}\), \(\mathrm{k+3}\), and \(\mathrm{k-1}\) form consecutive integers when rearranged. They fail to notice that while \(\mathrm{k-1}\) and \(\mathrm{k}\) are consecutive, \(\mathrm{k+3}\) creates a gap, breaking the consecutive pattern that guarantees divisibility by 3.

Faltering Point 3: Testing only one value and drawing conclusions
Students might test choice A with \(\mathrm{k = 7}\) (which gives 420, divisible by 3) and incorrectly conclude that choice A is always divisible by 3. They need to test multiple values to find a counterexample.

Errors while selecting the answer

Faltering Point 1: Selecting the first expression that works instead of the exception
After finding that choice E (consecutive integers) is always divisible by 3, students might select it as their answer, forgetting they need to find the expression that is NOT always divisible by 3.

Faltering Point 2: Doubting the counterexample
Even after correctly finding that \(\mathrm{k = 8}\) makes choice A equal to 616 (not divisible by 3), students might second-guess their arithmetic and choose a "safer" option like choice B, thinking they made a calculation error.

Alternate Solutions

Smart Numbers Approach

Step 1: Select strategic smart numbers
Since we need to test divisibility by 3, we'll choose values of \(\mathrm{k}\) that represent different remainders when divided by 3. This systematic approach will help us identify patterns:
• \(\mathrm{k = 7}\) (remainder 1 when divided by 3)
• \(\mathrm{k = 8}\) (remainder 2 when divided by 3)
• \(\mathrm{k = 9}\) (remainder 0 when divided by 3)

Step 2: Test each answer choice with \(\mathrm{k = 7}\)
A. \(\mathrm{k(k + 3)(k - 1) = 7(10)(6) = 420}\)
\(\mathrm{420 ÷ 3 = 140}\) ✓ (divisible)
B. \(\mathrm{3k^3 = 3(7^3) = 3(343) = 1,029}\)
Always divisible by 3 since it has factor 3 ✓
C. \(\mathrm{(k+1)(k+5)(k+6) = (8)(12)(13) = 1,248}\)
\(\mathrm{1,248 ÷ 3 = 416}\) ✓ (divisible)
D. \(\mathrm{(k+2)(k-2)(k+3) = (9)(5)(10) = 450}\)
\(\mathrm{450 ÷ 3 = 150}\) ✓ (divisible)
E. \(\mathrm{k(k+1)(k+2) = 7(8)(9) = 504}\)
\(\mathrm{504 ÷ 3 = 168}\) ✓ (divisible)

Step 3: Test each answer choice with \(\mathrm{k = 8}\)
A. \(\mathrm{k(k + 3)(k - 1) = 8(11)(7) = 616}\)
\(\mathrm{6+1+6 = 13}\), not divisible by 3 ✗
B. \(\mathrm{3k^3 = 3(8^3) = 3(512) = 1,536}\)
Always divisible by 3 ✓
C. \(\mathrm{(k+1)(k+5)(k+6) = (9)(13)(14) = 1,638}\)
\(\mathrm{1,638 ÷ 3 = 546}\) ✓ (divisible)
D. \(\mathrm{(k+2)(k-2)(k+3) = (10)(6)(11) = 660}\)
\(\mathrm{660 ÷ 3 = 220}\) ✓ (divisible)
E. \(\mathrm{k(k+1)(k+2) = 8(9)(10) = 720}\)
\(\mathrm{720 ÷ 3 = 240}\) ✓ (divisible)

Step 4: Confirm the pattern
Choice A fails the divisibility test with \(\mathrm{k = 8}\), making it the exception. Let's verify with \(\mathrm{k = 9}\):
A. \(\mathrm{k(k + 3)(k - 1) = 9(12)(8) = 864}\)
\(\mathrm{864 ÷ 3 = 288}\) ✓ (divisible)

Step 5: Identify the key insight
Choice A is the only expression that is not ALWAYS divisible by 3. When \(\mathrm{k ≡ 2 \pmod{3}}\), the factors \(\mathrm{k}\), \(\mathrm{k+3}\), and \(\mathrm{k-1}\) have remainders 2, 2, and 1 respectively when divided by 3, so their product has remainder \(\mathrm{2×2×1 = 4 ≡ 1 \pmod{3}}\), making it not divisible by 3.

Answer: A