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

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Intern
Joined: 26 Aug 2015
Posts: 38
Concentration: Strategy, Economics
GMAT 1: 570 Q40 V28
GMAT 2: 740 Q49 V41
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, 11:45
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67% (01:05) correct 33% (01:13) wrong based on 180 sessions

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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.
[Reveal] Spoiler: OA

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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, 12:01
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.
Intern
Joined: 26 Aug 2015
Posts: 38
Concentration: Strategy, Economics
GMAT 1: 570 Q40 V28
GMAT 2: 740 Q49 V41
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, 12:18
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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, 07:07
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(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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15 Jan 2018, 04:36
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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] 15 Jan 2018, 04:36
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