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

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

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New post Updated on: 04 Oct 2017, 03:10
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If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4

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Originally posted by vivek6199 on 04 Oct 2017, 03:08.
Last edited by Bunuel on 04 Oct 2017, 03:10, edited 1 time in total.
Renamed the topic.
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 04 Oct 2017, 03:13
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 04 Oct 2017, 03:44
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vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


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

Bunuel has offered the best algebraic approach.
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 06 Oct 2017, 10:58
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4



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
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 10 Oct 2017, 10:17
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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!
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 10 Oct 2017, 10:25
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zflodeen wrote:
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!


ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check below for more:
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Hope it helps.
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 10 Oct 2017, 10:37
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.
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 26 Jul 2018, 10:22
Bunuel wrote:
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


Option A: \(n^3 – 4n = n(n^2-4)=(n-2)n(n+2)\).

(n-2)n(n+2) is the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

Answer: A.


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 + \(a^2+b^2\)
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 26 Jul 2018, 10:35
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dave13 wrote:
Bunuel wrote:
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


Option A: \(n^3 – 4n = n(n^2-4)=(n-2)n(n+2)\).

(n-2)n(n+2) is the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

Answer: A.


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 + \(a^2+b^2\)


Because n^2 + 4 does not equal to (n + 2)(n - 2).
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If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 26 Jul 2018, 10:46
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dave13 wrote:
Bunuel wrote:
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


Option A: \(n^3 – 4n = n(n^2-4)=(n-2)n(n+2)\).

(n-2)n(n+2) is the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

Answer: A.


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 + \(a^2+b^2\)


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!
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 26 Jul 2018, 21:20
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vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


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.
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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 31 Oct 2018, 05:11
Bunuel wrote:
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


Option A: \(n^3 – 4n = n(n^2-4)=(n-2)n(n+2)\).

(n-2)n(n+2) is the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

Answer: A.


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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Re: If n is an integer, which of the following must be divisible by 3?  [#permalink]

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New post 31 Oct 2018, 05:26
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Mansoor50 wrote:
Bunuel wrote:
vivek6199 wrote:
If n is an integer, which of the following must be divisible by 3?

A) n^3 – 4n
B) n^3 + 4n
C) n^2 +1
D) n^2 -1
E) n^2 -4


Option A: \(n^3 – 4n = n(n^2-4)=(n-2)n(n+2)\).

(n-2)n(n+2) is the product of three consecutive odd or three consecutive even integers. In any case one of them must be a multiple of 3.

Answer: A.


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


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)
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Re: If n is an integer, which of the following must be divisible by 3?   [#permalink] 31 Oct 2018, 05:26
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