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If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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21 Jul 2016, 12:45
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If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following? a) 3 b) 5 c) 6 d) 8 e) 16 Question on NOVA's GMAT Math Prep Course.
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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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21 Jul 2016, 13:01
Ilomelin wrote: Question on NOVA's GMAT Math Prep Course.
if \(n\) is a positive integer and \((n+1)(n+3)\) is odd, then \((n+2)(n+4)\) must be a multiple of which one of the following?
a) 3
b) 5
c) 6
d) 8
e) 16 NOTE: When x is divisible by y means y is a factor/multiple of x.Based on this we'll take up the case. This is a must be case , given (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple . Now take 8 from the options : then 9*11 is odd , then 10*12 = 120 is even and 8 is factor of 120. Hence this is must be case. Now take 16 from options: then 17*19 is odd then 18*20 = 360 is even , here 16 is not factor of 360. This is not must be case. Hence option D is correct option.



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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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21 Jul 2016, 13:18
Ilomelin wrote: Question on NOVA's GMAT Math Prep Course.
if \(n\) is a positive integer and \((n+1)(n+3)\) is odd, then \((n+2)(n+4)\) must be a multiple of which one of the following?
a) 3
b) 5
c) 6
d) 8
e) 16 This question can be solved the following way: 1) Since \((n+1)(n+3)\) is odd, we can say that each term is odd. 2) \(n+1=\)odd then n must be even. 3) if \(n =\) even, then \(n = 2m\), where \(m\) is any constant. 4) \((2m+2)(2m+4) = 2(m+1)*2(m+2) = 4(m+1)(m+2)\) Since \(m+1\) and \(m+2\) are consecutive integers, one of them MUST be even and therefore have a factor of 2. \(4*2 = 8\), D is the answer.
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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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15 Oct 2016, 08:07
(n+1)(n+3) = odd So, n+1 and n+3 are odd So, n+1 = odd > n is even ; idem for n +3
so let s try the first values
n=2 > (2+2)*(2+4) = 2^3*3^1 n=4 > (4+2)*(4+4) = 2^4*3^1
(here, you cannot decide whether 6 or 8 is right answer, so let's continue)
n=6 > (6+2)*(6+4) = 2^4*5....
So only 8 MUST be a multiple of (n+2)*(n+4)



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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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20 Mar 2018, 20:02
Ilomelin wrote: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following?
a) 3 b) 5 c) 6 d) 8 e) 16
Question on NOVA's GMAT Math Prep Course. since (n + 1)(n+3) = odd, we can say with certainty that "n" is even now, if we assume "n= 0", then (n + 2)(n + 4) = 8, which is a multiple of 8 if we assume "n = 2", then (n + 2)(n + 4) = 24, which is a multiple of 8 thus, in both cases, "n" is a multiple of 8 = D the answer thanks



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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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26 Mar 2018, 00:32
gmatcracker2018 wrote: Ilomelin wrote: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following?
a) 3 b) 5 c) 6 d) 8 e) 16
Question on NOVA's GMAT Math Prep Course. since (n + 1)(n+3) = odd, we can say with certainty that "n" is even now, if we assume "n= 0", then(n + 2)(n + 4) = 8, which is a multiple of 8 if we assume "n = 2", then (n + 2)(n + 4) = 24, which is a multiple of 8 thus, in both cases, "n" is a multiple of 8 = D the answer thanks Question says n > 0.
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If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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Updated on: 26 Mar 2018, 03:05
Ilomelin wrote: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following?
a) 3 b) 5 c) 6 d) 8 e) 16
Question on NOVA's GMAT Math Prep Course. We need "n" as even, therefore take \(n = 2\) Then \((n+1)(n+3) = 3*5 = 15,\) and \((n+2)(n+4) = 4*6 = 24\) Therefore (B) and (E) are out. Take \("n" = 6\) \((n+2)(n+4) = 8*10 = 80\) Eliminate (A) and (C) are out Answer (D)
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Originally posted by QZ on 26 Mar 2018, 00:57.
Last edited by QZ on 26 Mar 2018, 03:05, edited 1 time in total.



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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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26 Mar 2018, 01:04
QZ wrote: gmatcracker2018 wrote: Ilomelin wrote: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following?
a) 3 b) 5 c) 6 d) 8 e) 16
Question on NOVA's GMAT Math Prep Course. since (n + 1)(n+3) = odd, we can say with certainty that "n" is even now, if we assume "n= 0", then(n + 2)(n + 4) = 8, which is a multiple of 8 if we assume "n = 2", then (n + 2)(n + 4) = 24, which is a multiple of 8 thus, in both cases, "n" is a multiple of 8 = D the answer thanks Question says n > 0. oh! sure, man "n" cannot be '0' you got it right: algebra is okay here thanks to note it + 1



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Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must [#permalink]
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26 Mar 2018, 01:06
QZ wrote: Ilomelin wrote: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must be a multiple of which one of the following?
a) 3 b) 5 c) 6 d) 8 e) 16
Question on NOVA's GMAT Math Prep Course. We need "n" as even, therefore take \(n = 2\) Then \((n+1)(n+3) = 3*5 = 15,\) and \((n+2)(n+4) = 4*6 = 24\) Therefore (B) and (D) are out. Take \("n" = 6\) \((n+2)(n+4) = 8*10 = 80\) Eliminate (A) Take \("n" = 8\) \((n+2)(n+4) = 10*12 = 120\) Eliminate (E) Answer (D) yes, its okay but I think algebra is quite safe here thanks




Re: If n is a positive integer and (n+1)(n+3) is odd, then (n+2)(n+4) must
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