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If \(\mathrm{n}\) is an integer greater than \(\mathrm{6}\), which of the following must be divisible by \(\mathrm{3}\)?
Let's start by understanding what we're looking for in plain English. We have five different expressions, each involving three numbers that depend on n. Our job is to find which expression will ALWAYS give us a result that can be divided evenly by 3, no matter what integer we choose for n (as long as n is greater than 6).
Think of it this way: if I give you any number for n - say \(\mathrm{n = 10}\), or \(\mathrm{n = 15}\), or \(\mathrm{n = 23}\) - whichever expression we choose must always produce a result that's divisible by 3. It's not enough for it to work sometimes; it must work every single time.
Process Skill: TRANSLATE
Here's a key insight: if we're multiplying several numbers together, the entire product will be divisible by 3 if at least one of those numbers is divisible by 3.
For example, if we have \(\mathrm{4 \times 7 \times 9}\), this product is divisible by 3 because 9 is divisible by 3. It doesn't matter that 4 and 7 aren't divisible by 3 - as long as one factor is divisible by 3, the whole product will be.
So our strategy becomes: look at each answer choice and see if we can guarantee that at least one of the three factors will always be divisible by 3, regardless of what n we choose.
Now here's the beautiful pattern that makes this problem solvable: among any three consecutive integers, exactly one of them must be divisible by 3.
Let me show you what I mean with some examples:
This always works because as we count 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ..., every third number is divisible by 3.
Now let's check our answer choices to see which ones give us three consecutive integers:
Choice A: \(\mathrm{n(n+1)(n-4)}\) → This gives us n, n+1, and n-4. Let's see: if \(\mathrm{n = 10}\), we get 10, 11, and 6. These aren't consecutive.
Wait, let me be more systematic. Let me rewrite each expression to see the three factors clearly:
Looking at Choice B more carefully: \(\mathrm{(n-1), n, (n+1)}\) - these ARE three consecutive integers!
But wait, let me double-check the original expression for Choice A: \(\mathrm{n(n+1)(n-4)}\)
Actually, let me rearrange this: we have n-4, and then we skip to n, n+1
Hmm, let me reconsider Choice A more carefully: \(\mathrm{n(n+1)(n-4)}\)
If I rearrange the factors in order: \(\mathrm{(n-4), n, (n+1)}\)
From n-4 to n is a gap of 4, and from n to n+1 is a gap of 1. So these aren't consecutive either.
Let me check if there's another pattern. Actually, let me look at this more systematically by testing a specific value.
Process Skill: INFER
Let me test each choice with \(\mathrm{n = 7}\) (since \(\mathrm{n > 6}\)):
Choice A: \(\mathrm{n(n+1)(n-4) = 7 \times 8 \times 3 = 168}\)
\(\mathrm{168 \div 3 = 56}\) ✓ (divisible by 3)
Choice B: \(\mathrm{n(n+2)(n-1) = 7 \times 9 \times 6 = 378}\)
\(\mathrm{378 \div 3 = 126}\) ✓ (divisible by 3)
Both work for \(\mathrm{n = 7}\)! Let me try \(\mathrm{n = 8}\):
Choice A: \(\mathrm{n(n+1)(n-4) = 8 \times 9 \times 4 = 288}\)
\(\mathrm{288 \div 3 = 96}\) ✓ (divisible by 3)
Choice B: \(\mathrm{n(n+2)(n-1) = 8 \times 10 \times 7 = 560}\)
\(\mathrm{560 \div 3 = 186.67...}\) ✗ (not divisible by 3)
Aha! Choice B doesn't always work. Let me try \(\mathrm{n = 10}\) for Choice A:
Choice A: \(\mathrm{n(n+1)(n-4) = 10 \times 11 \times 6 = 660}\)
\(\mathrm{660 \div 3 = 220}\) ✓ (divisible by 3)
Now I see the pattern for Choice A! Looking at the factors \(\mathrm{(n-4), n, (n+1)}\), I notice that these represent three numbers where we have consecutive integers among any group of 6 consecutive integers. Since among any 6 consecutive integers, exactly two will be divisible by 3, we're guaranteed to always have at least one multiple of 3 among our three factors.
More precisely: among the numbers \(\mathrm{(n-4), (n-3), (n-2), (n-1), n, (n+1)}\), exactly two are divisible by 3, so at least one of our three chosen numbers \(\mathrm{(n-4), n, (n+1)}\) must be divisible by 3.
The correct answer is A: \(\mathrm{n(n+1)(n-4)}\)
This expression is always divisible by 3 because among the three factors \(\mathrm{(n-4), n, and (n+1)}\), at least one must always be divisible by 3, regardless of the value of n. When we multiply any set of integers where at least one is divisible by 3, the entire product is divisible by 3.
1. Misunderstanding the requirement for "must be divisible"
Students often confuse "must be divisible by 3" with "could be divisible by 3." The question asks which expression is ALWAYS divisible by 3 for any integer \(\mathrm{n > 6}\), not which one works for some values of n. This leads students to test one or two values, see that multiple choices work, and get confused about which answer to select.
2. Focusing on consecutive integers incorrectly
Students may remember that "among any 3 consecutive integers, one is divisible by 3" and then incorrectly look for expressions with three consecutive terms like \(\mathrm{(n-1), n, (n+1)}\). This leads them to Choice B, missing that the actual factors in Choice A follow a different but equally valid divisibility pattern.
3. Overlooking the constraint n > 6
Students might ignore the given constraint and test small values like \(\mathrm{n = 1, 2, or 3}\), which could lead to different divisibility patterns or even negative factors that complicate the analysis. The constraint \(\mathrm{n > 6}\) ensures that all factors remain positive and the divisibility pattern holds consistently.
1. Arithmetic errors when testing values
When testing specific values of n, students often make calculation mistakes. For example, when checking Choice B with \(\mathrm{n = 8: n(n+2)(n-1) = 8 \times 10 \times 7 = 560}\), students might incorrectly calculate this as 540 or make errors when dividing by 3, leading them to think 560 is divisible by 3.
2. Incomplete testing of values
Students may test only one value of n (often \(\mathrm{n = 7}\)) and see that multiple choices work, then assume any of those choices is correct. They fail to test additional values systematically to eliminate incorrect options, missing that some expressions only work for specific values rather than all values.
3. Misidentifying factors in the expressions
Students may incorrectly rearrange or identify the three factors in each choice. For instance, in Choice A: \(\mathrm{n(n+1)(n-4)}\), they might not recognize that the three factors are n, (n+1), and (n-4), or they might incorrectly try to group terms together, leading to wrong analysis of the divisibility pattern.
1. Selecting based on partial verification
After finding that multiple choices work for their tested value (like \(\mathrm{n = 7}\)), students might select the first one that worked or the one that "looks simpler" rather than systematically eliminating wrong choices. They don't realize they need to find the choice that works for ALL values of \(\mathrm{n > 6}\).
2. Confusing sufficient testing with proof
Students might test 2-3 values, see that Choice A works for all of them, and select it without fully understanding WHY it always works. While this leads to the correct answer, they miss the underlying mathematical reasoning about divisibility patterns and might not be able to solve similar problems confidently.
Step 1: Choose strategic values for n
Since we need to test which expression is always divisible by 3 for any integer \(\mathrm{n > 6}\), let's choose values that represent different cases based on their remainder when divided by 3:
These three values cover all possible cases since any integer must have remainder 0, 1, or 2 when divided by 3.
Step 2: Test each answer choice with n = 7
Step 3: Test remaining choices with n = 8
We can eliminate C and D, so we test A, B, and E:
Step 4: Verify choice A with n = 9
Step 5: Understand why A always works
From our testing, only choice A consistently produces results divisible by 3. Looking at the pattern \(\mathrm{n(n+1)(n-4)}\), we can rewrite this as \(\mathrm{n(n+1)(n-4)}\). Notice that we can rearrange the factors as \(\mathrm{(n-4), n, (n+1)}\), which represents three consecutive integers when we order them properly. Since among any three consecutive integers exactly one must be divisible by 3, their product will always be divisible by 3.
Conclusion: The answer is A, confirmed through strategic testing with smart numbers representing all possible remainders when dividing by 3.