If n is an integer, which of the following must be divisible by 3?

Hey dave13

\((a^2 - 4) = (a^2 - 2^2) = (a + b)(a - b)\)

Let a = 5 -> Left hand side is \((5^2 - 2^2) = 25 - 4 = 21\) | The right-hand side is \((5 + 2)(5 - 2) = 7*3 = 21\)

But try it with your formula - you say \((a^2 + 4) = (a + 2)(a + 2)\)

Let a = 3 -> Left hand side is \((3^2 + 2^2) = (9 + 4) = 13\) | The right-hand side is \((3 + 2)(3 + 2) = 5*5 = 25\)

Since these are not equal - your formula must be wrong

Hope this clears your confusion.

I have merely expanded on Bunuel 's option!

Solved in in one minute, plug in n=2 and n=3.

Bunuel has offered the best algebraic approach.

Let’s simplify each answer choice:

A) n^3 – 4n

n(n^2 - 4) = n(n + 2)(n - 2) = (n - 2)(n)(n + 2)

We see that the expression above is a product of 3 consecutive even integers (if n is even) or the product of 3 consecutive odd integers (if n is odd). In either case, the product will always contain a prime factor of 3, so n^3 – 4n is always divisible by 3.

Alternate Solution:

If we take n = 1, we see that 1^3 + 4 = 5 is not a multiple of 3; thus, B cannot be the answer.

If we take n = 1, we see that 1^2 + 1 = 2 is not a multiple of 3; thus, C cannot be the answer.

If we take n = 3, we see that 3^2 - 1 = 8 is not a multiple of 3; thus, D cannot be the answer.

If we take n = 3, we see that n^2 - 4 = 5 is not a multiple of 3; thus, E cannot be the answer.

The only remaining choice is A.

Answer: A

I really struggled on this one and feel like I am missing something on this question. Isn't it possible for n to be zero as the question does not suggest otherwise? If you take the algebraic breakdown for A being (n-2)n(n+2), isn't it possible to be (-2)(0)(2)? I may be missing a math concept here, but I don't believe (-2)(0)(2) is divisible by 3.

Can someone help clear this up for me?

Thank you!

Because n^2 + 4 does not equal to (n + 2)(n - 2).

OA: A

The product of three consecutive integers is divisible by 3.
Three consecutive integers be \(n-1, n,n+1\)
Product of these three consecutive integers : \((n-1)(n)(n+1)=(n^2-1)n=n^3-n\)
\(3n\) is also divisible by 3.
\((n^3-n)-(3n)\) i.e \(n^3 – 4n\) would also be divisible by 3.

As far as divisibility by 3 is concerned,

(n - 2) is the same as (n + 1) because if (n - 2) is divisible by 3, then so is (n + 1) (which is just (n - 2 + 3)). If (n - 2) leaves a remainder of 1, so will (n + 1). If (n - 2) leaves a remainder of 2, so will (n + 1).

By the same concept, (n + 2) is the same as (n - 1)

So in effect, what we are looking at is this: (n - 1)*n*(n + 1)

I think this problem is wrongly classified as Distance/Rate problem. If someone could please change it

Bunuel, thank you for clearing that up! I was missing point #4 and suspected that might be the case. I'm trying to dredge up all these old math concepts and appreciate all the insight on these forum posts.

pushpitkc, if Bunuel says the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

why cant I say the same of option B ?

\(n^3 + 4n = n(n^2+4)=(n+2)n(n+2)\). here is the same product of three consecutive odd or three consecutive even integers.

Bunuel used formula \(a^2-b^2\) so I just changed the sign onto + \(a^2+b^2\)

how is (n-2)(n)(n+2) consecutive?

if we take (n-3)(n)(n+3), for n as 5 this gives us: 2 x 5 x 8 and as can be seen, is not divisible by 3. But this divisible by 3 if n is even

for (n-4)(n)(n+4) is divisible by 3 for even or odd n

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Edited the tags. Thank you.

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