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

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New post 21 Jul 2016, 12:45
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Question Stats:

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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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New post 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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New post 21 Jul 2016, 13:18
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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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New post 15 Oct 2016, 08:07
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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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New post 20 Mar 2018, 20:02
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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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New post 26 Mar 2018, 00:32
1
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
:cool:


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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New post 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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New post 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
:cool:


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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New post 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   [#permalink] 26 Mar 2018, 01:06
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