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If n is an integer, which of the following must be divisible by 3?
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Updated on: 04 Oct 2017, 02:10
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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, 02:08.
Last edited by Bunuel on 04 Oct 2017, 02:10, edited 1 time in total.
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Re: If n is an integer, which of the following must be divisible by 3?
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04 Oct 2017, 02:13




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Re: If n is an integer, which of the following must be divisible by 3?
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04 Oct 2017, 02:44
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?
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06 Oct 2017, 09:58
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?
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10 Oct 2017, 09:17
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 (n2)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?
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10 Oct 2017, 09:25
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 (n2)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: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: If n is an integer, which of the following must be divisible by 3?
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10 Oct 2017, 09:37
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?
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26 Jul 2018, 09:22
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: \(n^3 – 4n = n(n^24)=(n2)n(n+2)\). (n2)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 \(a^2b^2\) 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?
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26 Jul 2018, 09:35
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: \(n^3 – 4n = n(n^24)=(n2)n(n+2)\). (n2)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 \(a^2b^2\) 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?
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26 Jul 2018, 09:46
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: \(n^3 – 4n = n(n^24)=(n2)n(n+2)\). (n2)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 \(a^2b^2\) 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 righthand 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 righthand 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?
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26 Jul 2018, 20:20
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 \(n1, n,n+1\) Product of these three consecutive integers : \((n1)(n)(n+1)=(n^21)n=n^3n\) \(3n\) is also divisible by 3. \((n^3n)(3n)\) 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?
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31 Oct 2018, 04:11
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: \(n^3 – 4n = n(n^24)=(n2)n(n+2)\). (n2)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 (n2)(n)(n+2) consecutive? if we take (n3)(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 (n4)(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?
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31 Oct 2018, 04:26
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: \(n^3 – 4n = n(n^24)=(n2)n(n+2)\). (n2)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 (n2)(n)(n+2) consecutive? if we take (n3)(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 (n4)(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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